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Peter Borwein
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CMS Books in Mathematics
Ouvrages de mathematiques de la SMC
1 Herman/Kucera/Simsa Equations and Inequalities
2 Arnold Abelian Groups and Representations of Finite Partially Ordered Sets
3 Borwein/Lewis Convex Analysis and Nonlinear Optimization
4 Levin/Lubinsky Orthogonal Polynomials for Exponential Weights
5 Kane Reflection Groups and Invariant Theory
6 Phillips Two Millennia of Mathematics
7 Deutsch Best Approximations in Inner Product Spaces
8 Fabian et al. Functional Analysis and Infinite-Dimensional Geometry

Marian Fabian Petr Habala
Petr Hajek
Vicente Montesinos Santalucia
Jan Pelant Vaclav Zizler
Functional Analysis
and Infinite-Dimensional
Geometry
Springer

Marian Fabian
Czech Academy of Sciences
Mathematical Institute
AS CR, Zitna 25
115 67, Prague 1
Czech Republic
fabian@math.cas.cz
Petr Habala
Department of Mathematics
Faculty of Electrical Engineering
Czech Technical University
K301, FEL, CVUT
Technicka 2
166 27, Prague 6
Czech Republic
phabala@math.feld.cvut.cz
Petr Hajek
Czech Academy of Sciences
Mathematical Insitute
AS CR, Zitna 25
115 67, Prague 1
Czech Republic
Vicente Montesinos Santalucia
Department of Applied
Mathematics
Telecommunication Engineering
Faculty
Polytechnic University
of Valencia
46071 Valencia
Spain
vmontesi@mat.upv.es
Jan Pelant
Czech Academy of Sciences
Mathematical Institute
AS CR, Zitna 25
115 67, Prague 1
Czech Republic
pelant@math.cas.cz
Vaclav Zizler
Department of Mathematics
University of Alberta
Edmonton, Alberta
Canada T6G 2G1
Editors-in-Chief
Redacteurs-en-chef
Jonathan Borwein
Peter Borwein
Centre for Experimental and Constructive Mathematics
Department of Mathematics and Statistics
Simon Fraser University
Burnaby, British Columbia V5A 1S6
Canada
Mathematics Subject Classification B000): 46-01, 46Bxx
Library of Congress Cataloging-in-Publication Data
Functional analysis and infinite-dimensional geometry / Marian Fabian... [et al.].
p. cm. — (CMS books in mathematics ; 8)
Includes bibliographical references and index.
ISBN 0-387-95219-5 (alk. paper)
1. Functional analysis. 2. Banach spaces. I. Fabian, Marian J. II. Series.
QA320 .F793 2001
515'.1—dc21 00-053773
Printed on acid-free paper.
© 2001 Springer-Verlag New York, Inc.
All rights reserved. This work may not be translated or copied in whole or in part without the written permission
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Preface
Banach spaces provide a framework for linear and nonlinear functional
analysis, operator theory, abstract analysis, probability, optimization, and
other branches of mathematics. This book is intended as an introduction to
linear functional analysis and to some parts of infinite-dimensional Banach
space theory.
The first seven chapters are directed mainly to undergraduate and
graduate students. We have strived to make the text easily readable and as
self-contained as possible. In particular, we proved many basic facts that
are considered "folklore." An important part of the text is a large number
of exercises with detailed hints for their solution. They complement the
material in the chapters and contain many important results.
The last five chapters introduce the reader to selected topics in the theory
of Banach spaces related to smoothness and topology. This part of the book
is intended as an introduction to and a complement of existing books on the
subject ([Bea], [BeLi], [DGZ3], [Disl], [Dis2], [Fab], [JoL3], [LiT2], [Phe2],
[Woj]). Some material is presented here for the first time in a monograph
form. For further reading in this area, we recommend for instance [Gil],
[God4], [Gue], [JoL3], [Kec], [LjSo], [Neg], [MeNe], [Oxt], [RoJa], [Sem],
[Sin3], [Tal2], and [Vae].
The text is based on graduate courses taught at the University of Alberta
in Edmonton in the years 1984-1997. These courses were also taken by
many senior students in the Honors undergraduate program in Edmonton.
As a prerequisite, basic courses in calculus and linear algebra should be
enough. For the most part, Royden's book [Roy] should be sufficient.
The chapters are best read consecutively. However:

vi Preface
— Chapter 4 as well as the latter part of Chapter 3 (James boundaries)
can be omitted in the case of a more elementary functional analysis course.
Chapter 4 is used only marginally in Chapters 8-10.
— The spectral theory (Chapter 7) can be approached after the first two
chapters and the beginning of Chapter 3 are covered; it is not needed in
later chapters.
The book can serve as a textbook for the following types of courses in
functional analysis:
1. Graduate two-semester course: Chapters 1-9.
2. Graduate one-semester course: Chapters 1-3, 5, and 6 or 7.
3. Graduate one-semester advanced course: Chapters 8-10 or 11 and 12.
4. Undergraduate first course in functional analysis: Chapters 1-3 and a
part of Chapter 7.
5. Undergraduate second course in functional analysis: Chapters 4-6, 8,
and 10.
The first three chapters together with Chapter 7 can be used in service
courses for students of probability, physics, or engineering.
The principal part of the text was prepared at the Department of
Mathematics, University of Alberta in Edmonton. Each author spent some time
at this department. Habala and Hajek obtained their Ph.D. degrees there,
and Zizler was a faculty member there. We all thank this department for
excellent working conditions. We also thank our present home institutions
for enabling us to finalize the book. We are indebted to the grant agencies in
Canada, the Czech Republic, Germany, Spain, and the U.S. for supporting
our research in Banach space theory over the years.
We are grateful to our colleagues and students for many helpful
discussions. Our special thanks go to Jon Borwein, Robert Deville, Gilles
Godefroy, Jifi Jelinek, Kamil John, Lopez Pellicer, Jose Orihuela, Nicole
Tomczak-Jaegermann, Jon Vanderwerff, Dirk Werner, and John Whitfield.
We also thank those of our colleagues who allowed us to include some of
their recent unpublished results.
We thank Marion Benedict for her excellent typing of the first version of
the manuscript, and the staff of Springer-Verlag for their efficient work.
Above all, we are deeply indebted to our wives for their support and
encouragement.
We would be glad if this book inspired some young mathematicians
to choose Banach spaces as their field of interest. We wish the readers
a pleasant time spent over this book.
Prague, Czech Republic, Marian Fabian
and Valencia, Spain Petr Habala
Spring 2001 Petr Hajek
Vicente Montesinos Santalucia
Jan Pelant
Vaclav Zizler

Contents
Preface v
1 Basic Concepts in Banach Spaces 1
Holder and Minkowski inequalities, classical spaces C[0,1], ?p, Co,
MM],- • :••.•••; • • 2
Operators, quotient spaces, finite-dimensional spaces, Riesz's
lemma, separability 10
Hilbert spaces, orthonormal bases, li . 16
Exercises 21
2 Hahn-Banach and Banach Open Mapping Theorems 37
Hahn-Banach extension and separation theorems 38
Duals of classical spaces 44
Banach open mapping theorem, closed graph theorem, dual
operators 49
Exercises 52
3 Weak Topologies 63
Weak and weak star topology, Banach-Steinhaus uniform
boundedness principle, Alaoglu's and Goldstine's theorem,
reflexivity 64
Extreme points, Krein-Milman theorem, James boundary,
Ekeland's variational principle, Bishop-Phelps theorem . 76
Exercises 86

viii Contents
4 Locally Convex Spaces 107
Local bases, bounded sets, metrizability and normability,
finite-dimensional spaces, distributions 108
Bipolar theorem, Mackey topology 116
Caratheodory and Choquet representation; Banach-Dieudonne,
Eberlein-Smulian, Kaplansky theorems, and Banach-Stone
theorem 121
Exercises 130
5 Structure of Banach Spaces 137
Projections and complementability, Auerbach bases 137
Separable spaces as subspaces of C[0,1] and quotients of ?i,
Sobczyk's theorem, Schur's property of ?\ 140
Exercises 147
6 Schauder Bases 161
Shrinking and boundedly complete bases, reflexivity, Mazur's
basic sequence theorem, small perturbation lemma ... 165
Block basis sequences, Pelczyriski's decomposition method and
subspaces of ?pi Pitt's theorem, Khintchine's inequality and
subspaces of Lp 172
Unconditional bases, James's theorem on containment of ?\ and
Co, James's space J, Bessaga-Pelczyriski theorem .... 180
Markushevich bases in separable spaces, their extension property,
Johnson's and Plichko's result on ?^ 188
Exercises 191
7 Compact Operators on Banach Spaces 203
Compact and finite-rank operators, Fredholm operators,
Fredholm alternative 203
Eigenvalues, eigenspaces, spectrum, spectral decomposition . . 210
Spectral theory of compact self-adjoint and compact normal
operators 216
Banach's contraction principle, nonexpansive mappings, Ryll-
Nardzewski theorem, Brouwer's and Schauder's theorems,
invariant subspaces 227
Exercises 231
8 Differentiability of Norms 241
Smulian's dual test, Kadec's Frechet-smooth renorming of spaces
with separable dual, Frechet differentiability of convex
functions 243
More on extremal structure, Lindenstrauss's result on strongly
exposed points and norm-attaining operators 255
Exercises 263

Contents ix
9 Uniform Convexity 285
Uniform convexity and uniform smoothness, ?p spaces 285
Finite representability, local reflexivity, superreflexive spaces and
EnfiVs renorming, Kadec's and Gurarii-Gurarii-James
theorems 291
Exercises 305
10 Smoothness and Structure 313
Smooth and compact variational principles, sub differential,
StegalPs variational principle 314
Partitions of unity, smooth approximation 328
Lipschitz homeomorphisms, Aharoni's embeddings into Co,
Heinrich-Mankiewicz results on linearization of Lipschitz
maps 331
Homeomorphisms, Mazur's theorem on ?pi Kadec's theorem . . 336
Smoothness in ?p and Hilbert spaces 340
Countable James boundary and saturation by Co 343
Exercises 346
11 Weakly Compactly Generated Spaces 357
Projectional resolutions, injections into c0(r), Eberlein compacts,
embedding into a reflexive space, locally uniformly rotund
and smooth renormings 358
Weakly compact operators, Davis-Figiel-Johnson-Pelczyiiski
factorization, absolutely summing operators, Pietsch
factorization, Dunford-Pettis property 370
Quasicomplements 377
Exercises 379
12 Topics in Weak Topology 387
Eberlein compacts, metrizable subspaces 388
Uniform Eberlein compacts, scattered compacts 394
Weakly Lindelof spaces, property C 402
Corson compacts, weak pseudocompactness in Banach spaces,
(Bx, w) Polish 409
Exercises 417
References 431
Index 445

1
Basic Concepts in Banach Spaces
Most of the theory presented in this text is valid for both real and
complex scalar fields. When the proofs are similar, we formulate the theorems
without specifying the field over which we are working. When theorems are
not valid in both fields or their proofs are different, we specify the scalar
field in the formulation of a theorem. K denotes simultaneously the real
(R) or complex (C) scalar field. We use N for {1,2,...}. All topologies
are assumed to be Hausdorff. In particular, by a compact space we mean
a compact Hausdorff space.
Definition 1.1
A nonnegaiive function || • || on a vector space X is called a norm on X if
@ \\x\\ > 0 for every x E X,
(ii) \\x\\ = 0 if and only ifx = Q,
(Hi) \\Xx\\ — \X\ \\x\\ for every x E X and every scalar X,
(iv) \\x + y\\ < \\x\\ + \\y\\ for every x,y E X (the triangle inequality).
A vector space with a norm (X, || • ||) is called a normed linear space (or
just a normed space).
A Banach space is a normed linear space (X, || • ||) that is complete in the
canonical metric defined by p(x) y) — \\x — y\\ for x, y E X.
Note that the function p(x,y) = \\x — y\\ is indeed a metric on X. To
check the triangle inequality, we write
p(x,z) = \\x-z\\ = \\x-y + y-z\\ < \\x - y\\ + \\y - z\\ = p(x,y) + p(y,z).
All topological notions in Banach spaces refer to the canonical metric given
by the norm, unless stated otherwise. In situations where more than one

2 1. Basic Concepts in Banach Spaces
normed space is considered, we will sometimes use || • \\x to denote the
norm of X.
Let X = (X, || • ||) be a normed space. Bx = {x G X] \\x\\ < 1} denotes
the closed unit ball of X, and Sx — {# G X; \\x\\ = 1} denotes the unit
sphere of (X, || • ||). If M C X, then span(M) stands for the linear hull (or
span) of M, that is, the intersection of all linear subspaces of X containing
M. Equivalently, span(M) is the smallest (in the sense of inclusion) linear
subspace of X containing M. Similarly, span(M) stands for the closed linear
hull of M; the convex hull of M will be denoted by conv(M), and conv(M)
denotes the closed convex hull of M.
For subsets A, B of a vector space X and a scalar a, we also write A+B =
{a + b] a G A,b G B} and olA — {aa\ a G A}.
By a "subspace" we always mean a linear subspace. Let Y be a subspace
of a normed space (X, || • ||). By (Y, || • ||) we denote Y endowed with the
restriction of || • || to Y.
Fact 1.2
Let Y be a subspace of a Banach space X.Y is a Banach space if and only
if Y is closed in X.
PROOF: Assume that Y is closed. Consider a Cauchy sequence {t/n}?°=1 in
Y. Since the norm on Y is the restriction of the norm of X, the sequence is
Cauchy in X and therefore converges to some y G X. Because Y is closed,
y eY and yn -> y in Y.
The other direction is proved by a similar argument.
?
We will now turn to some examples of Banach spaces.
Definition 1.3
The space C[0,1] denotes the vector space of all scalar-valued continuous
functions on [0,1], endowed with the norm
H/lloo = sup{|/(*)|; * € [0,1]} = max{|/(f)|; t G [0,1]}.
We easily check that C[0,1] is a normed space. Consider a Cauchy
sequence {/n}?Li in C[0,1]. Because |A(*)-/*(*)I < 11/* -/HL tne sequence
{/n(t)}2°=1 is a Cauchy sequence for every t. Set f(t) = lim (/n(^))« It
n—>oo
remains to show that / is continuous and fn—*f uniformly. Given e > 0,
there is no such that |/n(^) — fm{t)\ < ? for every t G [0,1] and every
n, m > no. By fixing n > no and letting m —* oo, we get |/n(^) — f(t)\ < e
for every n > n0 and every t G [0,1]. Let *o G [0,1] and e > 0 be fixed.
Choose S > 0 so that |/no(^) ~ fn0(to)\ < ? whenever \t —1$\ < 6. Then
1/@ "/(Ml < l/(*)-/n.(<)l + l/no@-/no(<0)l + l/n.(*0)-/(MI
< 3? whenever \t — to\ < 6.

1. Basic Concepts in Banach Spaces 3
Therefore, / G C[0,1] and {/n} converges uniformly (hence in || • ||) to /,
which shows that C[0,1] is complete.
Analogously, the space C(K) of continuous scalar functions on a compact
space K, endowed with the supremum norm, forms a Banach space.
Definition 1.4
The space P^ denotes the n-dimensional vector space of all n-tuples of
scalars {that is, Rn or Cn), endowed with the supremum norm || • H^
defined for x — (xi,..., xn) G P^ by
H^Hoo = max{|&i|; i = 1,..., n}.
Note that P^ is a special case of a C(K) space, where K = {1,..., k}.
In order to introduce the class of ?p spaces for p < co, we need to prove
the following classical inequalities.
Theorem 1.5 (Holder inequality)
Let p, q > 1 be such that i + i = 1. Then for all a*, bk G K; k — 1,..., n,
we have
X>m < (f>r)* • (X>i«)''•
fc = l k = l jb = l
For p = 2, q = 2, the inequality is known as the Cauchy-Schwarz
inequality.
In the proof of Theorem 1.5, we will use the following statement.
Lemma 1.6
Let p,q>l be such that ± + i = 1. Then ab<^ + h-^- for all a, 6 > 0,
Proof: Consider the graph of the function y = x9, x > 0, and the areas
v4i of the region bounded by the curves y = xp~l, y — 0, x = a, and A2 of
the region bounded by the curves y — xp~l, a: = 0, y = b. Clearly, A\ —
f° xP^dx = ?. Since or = y1/^-1) = y«-\ we get A2 = J* y^dy = ?.
From the graph of the function y — xv~l we get ab < A\ + A2 = V + T*
?
An alternative proof is by checking on extrema of the function (p(a) —
?¦ + h- - ab for a fixed 6 > 0.
p q
Proof of Theorem 1.5: We may assume that a2-,6z- > 0 and neither all
a{ nor all b{ are zero. For k = 1,..., n, define
4* = — r and ?* = — r.
(?*/)* (ev)9

4 1. Basic Concepts in Banach Spaces
We note that ? Akp = ? Bfcg = 1. By Lemma 1.6, we have for k =
k = l fc = l
1,..., n that A^i?* < j^a^ + |#fcg- Summing up this inequality for k =
1,... , n, we get
n n n 1
which implies the desired inequality.
?
Theorem 1.7 (Minkowski inequality)
?etf p G [1, oo). TAen /or all ai^bk G K; & = 1,..., n} we have
(E i«*+h\p)' < (jr i«»p) * + Qr m*) v
fe = l fc = l fc = l
Proof: The statement is trivial for p = 1. If p G (l,oo), let </ G (l,oo)
be such that ^ + ^ = 1. We may assume that a;, 6f > 0. Using the Holder
inequality and the fact that (p — l)q = p, we obtain
Ek+6*)p = E(afc+6fc)p_1(afc+6fc)
= E(a*+6fc)p~la*+E(a*+6*)p_l6*
<
(E
(afc + 6fc)(p)?)?(E^P)?
+(E^+M(p-1)?)?(E^P)?
= (E(^+^)P)'(E^P)"
+(E(fl*+»*)p)i(iy)'-
Dividing by (?(a* + Mp)?> we Set
D
Definition 1.8
Let p G [l,oo). The space (J^ denotes the n-dimensional vector space Kn,
endowed with the norm defined for x = (#i,..., xn) G ip by
wip = (Enp)'-
2 = 1

1. Basic Concepts in Banach Spaces 5
By Minkowski's inequality, || • || is indeed a norm on X.
We suggest that the reader draw pictures of the unit balls of & for various
values of p. The unit ball of ?\ is the square with vertices ±ei, ±e2, where
t{ are the standard unit vectors, t\ = A,0) and t2 — @,1). The unit ball
of ?%, is the square with vertices (±ei ± e2). The unit ball of ?\ is the disk
of radius 1 centered at the origin.
The difference between ?J[ and P^ becomes apparent once we increase
the dimension. It is already apparent in three dimensions: The unit ball of
?%q is a cube, whereas the unit ball of ?\ is an octahedron. The unit ball of
?\ is a Euclidean ball.
Definition 1.9
Let p G [1, oo). The space ?p = ?P(N) denotes the vector space of all scalar-
valued sequences x = {xi}^l1 satisfying Yl\xi\p < °°> endowed with the
norm
OO _i
iwip = (Emp)?-
2 = 1
When a scalar sequence x — {xi]<^zl is considered an element of ?p, we
use the notation x = (a:,-). This applies to other sequence spaces defined
below.
To see that the definition is correct, we must show that if x = (#2), y =
(Vi) G ?p, then x + y G ?p and ||x + y\\p < \\x\\p + ||y||p.
For every n < m G N, by the Minkowski inequality we have:
71 1 71 1 7i i 771 1771 1
CE>+»rO * (EmOMEwO * (?mp)?+E>ip)?-
2 = 1 2 = 1 2'=1 2 = 1 2 = 1
By passing m —» oo, we see that for every n,
2 = 1 2 = 1 2 = 1
Letting n —> oo, we get x + y E ?p and the triangle inequality follows.
Let ? = (z;)^ be a sequence of scalars. We define the support of x by
supp(z) = {i; Xi^O}.
Definition 1.10
The space ?oo = ^oo(N) denotes the vector space of all bounded scalar-valued
sequences endowed with the norm defined for x — (xi) G ^oo by
II^Hoo = sup{|z2|; i G N}.
The space coo denotes the subspace of ^ consisting of all x = (x{) such
that supp(#) is finite.
The space c = c(N) denotes the subspace of ?& consisting of all x = (#2)
sucA 2/m2 lim(iCj) exists and is finite.

6 1. Basic Concepts in Banach Spaces
The space Co = co(N) denotes the subspace of l^ consisting of all x — (x{)
such that lim(xi) = 0.
i—»-oo
Note that cq is the closure of coo in ^oo • Note also that if x = (xi) belongs
to coo or Co, then ||#|| = max{|#;|; i G N}.
Proposition 1.11
(i) For p G [l,oo], the space ?p is a Banach space.
(ii) The spaces c and cq are closed subspaces of l^ and thus they are
Banach spaces.
(Hi) The space coo is not complete.
In the proof we will use the following lemma. Recall that a sequence {xn}
in a normed space X is called bounded if there is a constant C > 0 such
that ||#n|| < C for every n G N.
Lemma 1.12
Let X be a normed space. If a sequence {xn}'^L1 C X is Cauchy, then it is
bounded in X.
Proof: Using the Cauchy property of {xn}, we find n0 G N such that
|| < 1 for every n > n0. Then ||jcn|| < \\xn - xno\\ + ||&n0|| <
1 + H^noll f°r every n> uq. Therefore, for every n G N,
IMI < max{||a?i||, \\x2\\,..., |kn0-i||, IKoll + !}•
D
Proof of Proposition 1.11:
(i): It is easily checked, using a method similar to that for C[0,1], that
^oo is a Banach space. Now consider p G [1, oo).
Let {xk}<^L1 be a Cauchy sequence in ?pj where xk = (xk). Given e > 0,
find A?o such that
OO l_
(E\'i-^y^e w
for every fc, / > &o- In particular, \x\ — x\\ < e for every kJ>ko and i G N,
hence the sequence {xf }j?=1 converges to some x\ for every i G N.
Put # = (x{). We will show that i E V By Lemma 1.12, there
is a constant C > 0 such that (E kflP] < C f°r every &• There-
f°re lEI^Pj ^ C ^or a^ ft,? G N. By letting Ar —> oo, we get
(n \ i / oo \ i-
E NPj < C for every n G N. Therefore, (E NP J < C and xGV
2 = 1 ' \ = 1 '
We will now show that xk -+ x in ?p. Given e > 0, we let / —» oo in (*)
(n \ —
J^ \xk — Xi\pj < e for every n G N and every & > &o- We let
«•—1 '
\*=i

1. Basic Concepts in Banach Spaces 7
n ¦—> oo to obtain
fc^-x^Y = \\xk-x\\p<e
2 = 1
for every k > fco- Therefore a?fc —> x in ^p.
(ii): Let xk — (a;*) E c and xk —>¦ a? in ^oo. For I; G N, we denote
Ik = lim (a?*). We will prove that lim D) exists and is finite by showing
«—»>oo fc —*-oo
that {/n}fcLi is Cauchy. Given ? > 0, let no be such that \\xn — a?m||oo < e
for all n, ra > n0. Thus, \xf — x™\ < e for every i E N and every n,m> no-
Fixing n,ra > no and letting i —> oo, we get |/m — /n| < ?. Therefore
lim (/fc) = / exists and is finite.
k—>oo
Since a^ —» a: = (a^) in ^x,, we have lim (arf) = X{ for all i E N. We
fc—»-oo
will show that lim (xi) = /, that is, ? E c. Given ? > 0, we find no so that
i—>-oo
||^fc — #||oo < e and \lk — l\ < e for k > no, in particular |x* — ar;|oo < e.
Then fix z'o so that |a??° - /no| < e for z > z0. We have for i > z'o,
|x. _ /| < \Xi - ^o| + l^o _ /nj + l^ _ /( < 3e
Thus lim(a?z) = / and x E c. This shows that c is closed in i^.
Similarly, we show that cq is a closed subspace of-?00. By Fact 1.2, the
spaces c and Co are Banach spaces.
(iii): By Fact 1.2, it is enough to show that coo is not closed in cq. To
this end, consider x E c$ defined by x — (a?f), where X{ — i for every
i, and xn — (l, |,..., ^, 0,...). Then xn E coo and a;n —> a? in Co since
||xn — a?||oo = ^~ —* 0 as n —> 00. However, a? {? coo-
D
More generally, for an abstract set V we introduce spaces ?p(T) and co(r).
Let p E [1,00). The space ip(T) consists of all functions /: T —+ K such that
]C I/(t)Ip < °°5 w^n the norm ( ]T) |/(t)|p) ?, where the sum is defined
by
5^ |/G)|p - sup{? |/G) |p; F a finite subset of r}.
The space ?00 (r) consists of all bounded functions /: T —> K and is
endowed with the supremum norm || • U^. Its subspace co(r) consists of
all functions / E 4>o(r) such that the set {7; |/G)| > e} is finite for
every e > 0. We also consider the space coo(r) of all functions / E loo (r)
whose support supp(/) = {7; f(j) / 0} is finite. Note that co(r) = coo(r)
(closure in ^(r))*.
Similarly as before, we show that ip(T) and Co(r) are Banach spaces.

8 1. Basic Concepts in Banach Spaces
Definition 1.13
Let p G [l,oo). The space Lp = Lp[0,l] denotes the vector space of all
classes of Lebesgue-measurable scalar functions f defined almost everywhere
on [0,1] (we identify functions that are equal almost everywhere) such that
L \f(t)\p dt < oo; endowed with the norm
i/ii,
= (?\f(tWdt)
We must check that Lp is a vector space. Indeed,
\f(t)+9(t)\p < ^maxd/WI, \g(t)\Y = 2Pmax(\f(t)f,\9(t)\p)
< 2P(\f(t)\P + \g(t)f),
so fQ \f + g\p dt < oo and (f + g) G Lp whenever f,g G Lp. Similarly,
ocf G Lp for every a G K and / G Lp.
The triangle inequality for || • || follows from Minkowski's inequality.
Theorem 1.14
Lfp G [l,oo); then Lp is a Banach space.
In the proof, we will use the following lemma. We say that a series ]>3 xi
in a Banach space is absolutely convergent if ?3 Ik* 11 < oo. Recall that J2 xi
n
is called convergent if the sequence sn = 53 xi ls convergent in X.
t=i
Lemma 1.15
A normed space X is a Banach space if and only if every absolutely
convergent series in X is convergent.
Proof: Assume that X is a Banach space and ?3 \\xn\\ < °°- Then, given
n
e > 0, there is no G N such that 53 ||a?,-|| < e for every n > k > uq. Thus
fc+i
n || n it n
for sn = 53 xi we nave Ikn -«]b|| = 53 ** ^ 53 IK II < 6- Therefore, the
2 = 1
fe+i
Jfe+1
sequence {sn}?Li is Cauchy and thus convergent in X, so every absolutely
convergent series in a Banach space is convergent.
Now assume that the condition on absolute convergence is true, and let
{xn}^Ll be a Cauchy sequence in X. First, we show that {xn}^=1 has a
convergent subsequence. To this end, choose a subsequence n\ < ri2 < ...
such that \\xnk — xni\\ < T~h for / > k. This is done by induction: first
choose n\ such that ||xni — xi\\ < 2 for every / > rt\. Then choose n^
such that ||arn2 — #/|| < 2~2 for every / > n2, etc. Put xno = 0 and set
nk
Vk = Xnk - xnk_1. We have xnk = 53 S/t, where ||yjb|| < 21-* for every
*=i
Ar > 2. Because 5311^11 < °°j ky our assumption we get that 53 2/* is
convergent and hence its partial sums xnk form a convergent sequence.

1. Basic Concepts in Banach Spaces 9
To show that the whole sequence {xn}%Li converges to the same element
is standard. This concludes the proof of Lemma 1.15.
?
Proof of Theorem 1.14: By Lemma 1.15, we must show that ]T/n is
a convergent series in Lp whenever fn G Lp satisfy ^ || fn\\p < oo. Denote
oo n
M = E II/»IIp- For n € N, we define functions gn(t) = ? |/t(t)|. Then
n=l ifc=l
\\9n\\P < t II \M II, < M, so fi \gn\P dt = So 9Pn dt<MP.
Jb = l
We have gn+i(t) > gn(t) for every n G N and t G [0,1]. Thus the limit
g(t) = lim (#nC0) exists (finite or +oo) for every t G [0,1]. Since gn > 0,
n—>oo
by Fatou's lemma J* gp dt < liminf (JJ,1 gpn dt) < Mp.
Therefore gp G ?i[0,1], so g(t) < oo on a set S of measure 1 in [0,1]. For
t G 5 we have J^ \fk(t)\ = g(t) < oo. Thus the sum s(t) = ^2 fk(t) is finite.
Define s(t) = +oo for t ? S. We must show that s = ^ /& in Lp. Denote
n
sn(t) = ]T /ik@- Then s = lim (sn) almost everywhere (on 5 at least),
so it is a measurable function on [0,1]. We also have |sn(^)| < 9n(t) < #B)
for every t G 5. Therefore |s(J)| < #(?) almost everywhere on [0,1], which
implies that f \s(t)\p dt < oo and hence s G Lp.
Finally, we have
\sn(t) - s(t)\r < 2? max(|sn(*)F\ \s(t)\p) < 2V(*)
almost everywhere on [0,1], f2pgp dt < oo, and \sn(t) — s(t)\ —» 0 almost
everywhere. By the Lebesgue dominated convergence theorem,
rl
/
JO
D
/o
This means that \\sn — s\\p —» 0, that is, ?^/fc = s in Lp.
Let / be a measurable function on [0,1]. We define
esssup(/) = inf(sup{/(*); teN})}
where the infimum is taken over all measurable subsets AT of [0,1] of
Lebesgue measure one. One can also use other equivalent definitions, for
instance
esssup(/) = inf{a; \{t G [0,1]; f(t) > a}\ = 0},
where \A\ denotes the Lebesgue measure of A. Clearly, / < esssup(/)
outside a set of measure zero.
Definition 1.16
The space Lqo — -^oo[0,l] denotes the vector space of all classes of

10 1. Basic Concepts in Banach Spaces
measurable functions f that are "essentially bounded" (i.e., such that
esssup(|/|) < oo), endowed with the norm ||/||oo = esssup(|/|).
It is easy to check that || • H^ is a norm on the space Loo- We will show
that it is a Banach space. First, observe that {/n}?Li being a Cauchy
sequence in L^ implies that lim (fn — /m) = 0 uniformly on a set of
m,n-+ oo
measure 1. Indeed, for k, I E N, put
Zt,i = {te [o, i]; \fk(t) - ft(t)\ > \\fk - /dioo}.
On [0,1]\(J Zk i, the functions {/n} form a uniformly Cauchy and therefore
uniformly convergent sequence. From this and from the completeness of the
supremum norm we obtain that Lqq is a Banach space.
Given a measure space (fi,/i) and p E [l,oo], the space Lp(fi,//) (also
denoted Lp(pi)) can be introduced in a similar fashion ([DiUh]).
We now begin an investigation of linear mappings between normed
spaces. Recall that a map T:(P,p) —+ (Q,q) between metric spaces is
called Lipschitz if there is C > 0 such that g(T(x),T(y)) < p(x,y) for
all xyy E P. In the case of normed spaces P,Q, this inequality becomes
\\T(x) — T(y)\\Q < C\\x — y\\p. Note that every Lipschitz map is continuous.
Proposition 1.17
Let {X,\\'\\x) and (Y, || • ||y) be normed spaces, and let T be a linear
mapping from X into Y. The following are equivalent:
(i) T is continuous on X.
(ii) T is continuous at the origin.
(Hi) There is C > 0 such that \\T(x)\\Y < C\\x\\x for every x E X.
(iv) T is Lipschitz.
(v) T(Bx) is a bounded set in Y.
Recall that a subset M of a Banach space X is called bounded if there is
C such that ||z|| < C for x E M.
Proof: (i) <=> (ii) follows from the linearity of T. (iii) <=$¦ (iv) follows
similarly since \\T(x) - T(y)\\Y = \\T(x - y)\\Y < C\\x - y\\x.
If (ii) is true, then, given e > 0, there is 6 > 0 such that ||T(e)||y < e
whenever \\x\\x < 6. If x E X, x -? 0, then ll^ra^llx = 6 and thus
\\T(S^)\\y < e. Therefore, ||T(a:)||y < e/6\\x\\x for every xeX, which
shows (iii). On the other hand, (iii) clearly implies (ii) with 6 = e/C.
Assuming (iii), we obtain that ||T(x)||y < C whenever x E Bx, so T(BX)
is bounded. On the other hand, if T(BX) is bounded and, say, ||T(a:)||y < C
for every x E BX) then, for every x E X, x ^ 0, we have ||^(jrafc)l|y ^ Cj
so ||T(^)||y < C||#||x- This shows the equivalence of (iii) and (v).
D

1. Basic Concepts in Banach Spaces 11
Definition 1.18
Lei X, Y be normed spaces, and let T be a linear map from X into Y. T is
called a bounded linear operator ifT(Bx) is bounded in Y.
We define the operator norm ofT by
\\T\\ = sup{||r(x)||y; xeBx}.
The space B(XyY) denotes the vector space of all bounded linear operators
from X into Y, endowed with the operator norm. In the case of X = Y,
weputB(X)=B(X,X).
By Proposition 1.17, operators from B(X, Y) are precisely the continuous
linear operators from X into Y. It is easy to check that B(X) Y) is indeed a
normed linear space. Note that ||T|| is the smallest number C that satisfies
(iii) in Proposition 1.17, and if T\X —> Y is a bounded linear operator,
then the kernel Ker(T) = {x ? X] T(x) = 0} is a closed subspace of X.
Proposition 1.19
Let X,Y be normed linear spaces. IfY is a Banach space, then B(X,Y) is
also a Banach space.
Proof: The proof of completeness is similar to that for the space C[0,1].
In particular, if Tn are linear and Tn(x) —> T(x) for all x ? X, then T must
be linear.
?
Operators in B(X, K) are called continuous linear functionals. We then
have H/ll = sup{|/(s)|; * € Bx} for / E B(X,K).
An operator T E B(X, Y) is called a linear isomorphism (or just an
isomorphism) if T is a bijection (one-to-one and onto Y) and T_1 E B(Y,X).
Note that the inverse map of a linear bijection is also a linear map.
Normed spaces X) Y are called linearly isomorphic (or just isomorphic)
if there is a linear isomorphism T of X onto Y. It is easy to see that an
isomorphism T carries Cauchy (convergent) sequences onto Cauchy
(convergent) sequences. Therefore, if X) Y are isomorphic normed spaces and
X is a Banach space, then Y is a Banach space as well.
An operator T E B(X,Y) is called a (linear) isomorphism from X into
Y if T is an isomorphism from X onto a subset T(X) of Y. Clearly, T(X)
is a subspace of Y, and since it is a complete space, it must be closed in Y.
An operator T E B(X,Y) is called a (linear) isomeiry\i\\T(x)\\Y = ||^||x
for every x E X. Spaces X,Y are called (linearly) isometric if there exists
a linear isometry T of X onto Y.
Let a vector space X be endowed with two norms, denote X\ — (X, || • ||x)
and X2 — (X, || • ||2). Norms || • ||x and || • ||2 are called equivalent if the
formal identity mapping Ix: X —> X defined by Ix (x) = x is an isomorphism
between the spaces X\ and X2, that is, if there exist constants c, C > 0
such that c\\x\\2 < \\x\\i < C||a?|J2 for every x ? X.

12 1. Basic Concepts in Banach Spaces
For a pair X, Y of normed spaces, we introduce the normed space X 0 Y
called a direct [topological) sum of X and Y, that consists of all ordered
pairs {x,y), x G X, y ? Y, endowed with the norm ||(z,y)|| = lkl|x + ||2/||y.
X and Y are isometric to subspaces {(?,0); x ? X} and {@,y); yG^}
of X 0 Y. If X and Y are Banach spaces, then so is X 0 Y.
Let Y be a closed subspace of a normed space X. For x G X, we consider
the coset ? relative to Y,
^{^1; (* - 2?) G Y} = {* + y; y G Y}.
The space X/Y = {?; z G X} of all cosets, together with the addition
and scalar multiplication defined by x -f y = x+y and Xx — Az, is clearly
a vector space. It is easy to check that ||?|| = inf{||y||; y G x} turns X/Y
into a normed space. Indeed, for any z ? x we have ||?|| = inf {||z — y\\] y G
Y} = dist(z, Y). Therefore, ? = 6 if and only if x G Y, because Y is closed.
If Y is a subspace of X, then dist(o:ar,Y) — |a| dist(z, Y). Therefore
||A? || = |A| ||?||. The triangle inequality follows since if xi, x<i are in X and
2/ij 2/2 are in Y, then
11*1 + *2 ~ (j/1 + 2/2)|| < |kl — 3/xH "+- ||^2 - 2/2||.
Therefore dist(xi + x2i Y) < dist(a?i, Y) + dist(a?2, Y).
Definition 1.20
Let Y be a closed subspace of a normed space X. The space X/Y endowed
with the canonical norm \\x\\ — inf{||#||; x ? x} is called the quotient space
of X with respect to Y.
Proposition 1.21
Let Y be a closed subspace of a Banach space X. Then X/Y is a Banach
space.
Proof: Assume that {i^}^! is Cauchy in X/Y. Choose a subsequence
{nfe}fcLi sucn tnat ll^nfc - ?nfc+i|| < 2~fc. Then choose xnk G xnk such that
\\xnk ~ xnk+i\\ < 2~fe. This choice is possible since for arbitrary xnk G xnk,
the distance of xnk to xnk+1 is less than 2~k. It is standard to check that
{xnk}<^Ll is Cauchy in X and therefore xnk —» x in X. Since \\xnk — x\\ <
\\xnk — x\\ for every &, we get that xnk —» x. Consequently, xn —» x in X/Y.
D
It is easy to check that (X 0 Y)/X is isomorphic to Y and (X 0 Y)/Y
is isomorphic to X. However, if Y is a closed subspace of X, then X may
not be isomorphic to Y 0 (X/Y) (see Chapter 12, the space D).
Proposition 1.22
Let X be a vector space. If X is finite-dimensional, then any two norms on
X are equivalent.
In particular, all finite-dimensional normed spaces are Banach spaces, and
every normed space of dimension n is isomorphic to t7^.

1. Basic Concepts in Banach Spaces 13
Consequently, if X is a Banach space and Y is a finite-dimensional
subspace of X, then Y is closed in X by Fact 1.2.
Proof: Let {ei,..., en} be an algebraic basis of X. We introduce a new
norm || • \\x on X by ||#||i = ^ |Aj| for x = X^*e*'- To cneck the triangle
inequality, for x = Yl ^*e* and V — J2@iei we write
ii*+»iii = Eia«+a-i<Eia'I+Ei^i = nii+nii-
We will show that arbitrary norm || • || on X is a Lipschitz function on
(X, || • HJ. Indeed, if x = J2^iei and V = ]C$e*> tnen
Ik-2/11 - |E(A'-^.-|<ElA»-AINI
< max{||ei||}^|Ai-ft| = max{||ef||} - ||ar — 3/||i.
Therefore, |||*|| - ||y||| < \\x - y|| < max{||c,-||}||x - 2/||i.
We note that Si = {x E X; ||a?||i = 1} is compact in (X, || • 1^). Indeed,
let xk E Su xk = E \*e* for ^N.We have ? |A*| = 1 for every jfc, and
*=1 2 = 1
thus {\f }^CL1 is bounded for every i. Let a subsequence {Ar/}g1 be such
that A2* —> Aj for every i. Then ^ |A2-* — At-| —> 0 as / —> oo, so we have
2 = 1
n
a?*1 —> a: as / —» oo, where x = X^A^e,-. Since ]P |Af*| = 1 for every /, we
2 = 1
have E |AZ| = 1 and thus x E Si.
Since || • || is continuous on the compact set Si, there exist constants
c > 0 and d > 0 such that c < IITT^l—11 < ^ ^or every nonzero # E X. From
the latter inequality we have c||#||i < ||ar|| < d\\x\\i for every x E X, so || • ||
is equivalent "to f| - ||x- Consequently, all norms are equivalent.
If X is an n-dimensional vector space and T is a linear bijection from X
onto ^2> we can define a norm || • ||2 on X by ||z||2 = ||T(o;)||^». Then T is
an isometry of (X, || • ||2) onto ?% and || • ||2 is an equivalent norm.
n
Similarly, we can show that ||(^,y)|| = (||a?||x + IMIy)* *s an equivalent
renorming of X 0 Y for every p > 1. Such a renormed space is denoted
To characterize finite-dimensional normed spaces, we need the following
statement. Recall that a subspace Y of a vector space X is called proper if
Y ?X.
Lemma 1.23 (Riesz)
Let X be a normed space. IfY is a proper closed subspace of X, then for
every e > 0 there is x E Sx such that dist(#, Y) > 1 — e.

14 1. Basic Concepts in Banach Spaces
Proof: Choose an arbitrary element z ? X/Y satisfying 1 > ||i|| > 1 — e.
Pick any z ? z, \\z\\ < 1, and set x = 2/||z||.
dist(*,y) .= dist(z,y)/||z|| - ||*||/||z|| > p|| > 1 - e.
D
Theorem 1.24
Let X be a normed space. X is finite-dimensional if and only if the unit
ball Bx of X is compact.
Proof: If X is finite-dimensional, then Bx is compact by the proof of
Proposition 1.22.
If X is infinite-dimensional, by Lemma 1.23 we find by induction an
infinite sequence xn G Sx such that dist(a?n,span{^i,..., #n_i}) > |.
Thus dist(xn, xm) > \ for all n ^ ra, and therefore {xn} does not have any
convergent subsequence. Hence, Bx is not compact.
?
Recall that a subset M of a topological space X is called dense in X if
M, the closure of M in X, is equal to X.
Definition 1.25
Let X be a normed linear space. We say that X is separable if there exists
a sequence {xi}fl1 in X that is dense in X.
Proposition 1.26
@ If P ? [1>°°)> thtn the space ?p is separable,
(ii) The spaces c and cq are separable.
(Hi) The space t^ is not separable.
Proof: (i): Consider in ?p the family T formed by all finitely supported
vectors with rational coefficients. Then T is countable. We will show that
T is dense in ?p. Given x ? ?p and e > 0, choose n0 G N such that
oo eP
^2 \xi\p < ~zr and tnen find rational numbers t*i, r2,..., rno_i such that
i=n0 ^
eP
\*i-ri\p < -—for i- l,...,n0- 1. Then s = (rl3 r2j..., rno_i, 0,...) is
in T and
no — l oo no —1 p p
ik-^=Ei^-^ip+ENp<E|r + V<?P-
Therefore, T is dense in ?p.
(ii): The proof of the separability of cq is similar to (i). The case of
the space c needs only one adjust ment-namely, we define T as the family
of vectors with rational coefficients such that the vectors are eventually
constant.

1. Basic Concepts in Banach Spaces 15
(iii): Assume that ?oq is separable, and let V be a dense countable set
in ^oo. The cardinal number of the family T of all subsets of N is the
cardinality of the continuum c; in particular, it is uncountable. For every
FEf, let xf denote the characteristic function of F in N. If i<\, F2 E T
and Fi / F2, then ||xfx - Xf2||oo > |XFi(n) - XF2{n)\ = 1 for some
n E Fi \ F2 or n E T2 \ T\. For each F E T, let ^ G D be chosen
such that ||xf - dF\\ < |. If F1 / F2) then ||dFl - dF2||oo > \. Indeed, if
we had \\dpY - dp2||oo < \, then
\\XF1 ~ XF2\\oo < WXF, - c/fJIoo + \\dFl ~ dF2\\oo + Ife " XF2|| < | < 1,
a contradiction. Therefore, the map F \-+ dp is one-to-one and maps an
uncountable set into a countable set, which is a contradiction. Thus, l^ is
not separable.
?
Similarly, we prove that co(T) and ^>(r), p E [l,oo], are nonseparable
for T uncountable.
Proposition 1.27
(i) The space C[0,1] is separable.
(ii) Ifp E [l,oo); then Lp is separable.
(iii) The space L^ is not separable.
Proof: We only discuss the real case (real-valued functions). In the
complex case, we consider real and imaginary parts of functions separately in
order to reduce the problem to the real case.
(i): The collection V of polynomials on [0,1] forms an algebra in C[0,1]
that separates the points of [0,1] and contains a constant function.
Therefore, the closure V in C[0,1] is C[0,1] by the Stone-Weierstrass theorem
(see, e.g., [Roy]). Since the countable set of polynomials on [0,1] with
rational coefficients is dense in V, we obtain that C[0,1] is separable.
(ii): It is a classical fact that C[0,1] is dense in Lp (see, e.g., [Roy]).
Therefore, the real case of (ii) follows from (i).
(iii): Consider the functions ft = X[o,t], t ? [0,1], where X[o,t] is the
characteristic function of [0,i\. Then \\ft — ff\\ = 1 if t / t', and a similar
argument as in the proof of Proposition 1.26 gives that L^ is not separable.
?
Proposition 1.28
B(?2) contains an isometric copy of ?oq and thus it is not separable.
Proof: Define a map ip from l^ into B(?2) as follows. If (a2) E ^ooj let
(p((di)) be the bounded linear operator from ?2 into ?2 defined for (x{)i E ?2
by (p((ai)):(xi)i »-» (a{Xi)i.
We claim that <p is a linear isometry from i^ into B(?2). It is enough to
check that if IKa*)!^ = 1, then the operator cp((a()) has norm 1 in B(?2).

16 1. Basic Concepts in Banach Spaces
First, note that if ||x||2 = (? |a?*|2)* = 1, then
||(a^)||2 - E|a,-|2|x,.|2)* < IKa.OllcoEkil2)*
and thus the operator ^>((a«)) nas norm at most 1. On the other hand, given
e > 0, choose n0 such that |ano| > l—e. Then ^((a0)(eno) = an0en0, which
has norm |ano|. Letting e —» 0, we obtain ||^((a*))|| = 1.
?
Hilbert Spaces
Recall that an inner product (or a scalar product or a dot product) on a
vector space X is a scalar-valued function (y)onlxl such that
A) for every y E X, the function x i-> (jd, y) is linear;
B) (z,y) = (y,x), where the bar denotes the complex conjugation;
C) (x, x) > 0 for every x E X\
D) (#, a?) = 0 if and only if x = 0.
Note that by A), @, y) = 0 for any |/Gl; hence also (y, 0) = 0 by B).
Theorem 1.29 (Cauchy-Schwarz inequality)
Let (x,y) be an inner product on a vector space X.
{%) For x, y G X, we have |(a?,y)| < y/(xix)y/(y)y).
{it) The function \\x\\ = y/(x,x) is a norm on X.
Proof: (i): If (y, y) — 0, we have y = 0 and the inequality is satisfied.
Assume that (y, y) > 0. Then
n^ / (g>2/) (gJ/) ^ / n l(g»2/)|2
°^lx-(^y,!B-(^y)=(x,x)-n^-'
and the statement follows.
(ii): We will check the triangle inequality. For x,y E X, we have
Ik + V\\2 = (x + y,x + y) = (x, x) + (y, y) + (a:, y) + (y, a?)
= (a:, x) + (y, y) + 2 Re(ar, y) < (x, x) + (y, y) + 2|(z, y)\
< (*,») + (y, 2/) + W(x> xW(y> v)
= (^/(^+y(^J-(IHI + ||y||J-
?
One immediate consequence of Theorem 1.29 is that (•, •) is a
continuous mapping from (X, \\ • ||) x (X, || • ||) to the scalar field. In particular,
it implies that for a fixed vector y E X, x ¦-+ (a?, y) is a continuous linear
functional on X.

1. Basic Concepts in Banach Spaces 17
Definition 1.30
A Banach space H is called a Hilbert space if there is an inner product
(•, •) on H such that \\x\\ = y/(x,x) for every x ? H.
It is straightforward to check that the norm || • || on a Hilbert space H
satisfies the parallelogram equality] namely, for every x,y ? H, we have
\\x + y\\2 + \\x-y\\2=2\\x\\2 + 2\\y\\2.
We also have the polarization identity:
(x,y)=\(\\x + y\\2-\\x-y\f)
in the real case, and
(*, y) = i (Ik + wll2 - II* - yll2 + *1I* + iytf - *1I* - iyf)
in the complex case.
On the other hand, assume that a norm || • || on a Banach space X
satisfies the parallelogram equality. If we define (x>y) by the preceding
equations, it turns out to be an inner product (Exercise 1.40) and ||#||2 =
(x,x). Thus, X is a Hilbert space.
Therefore, a Banach space X is a Hilbert space if and only if every two-
dimensional subspace of X is a Hilbert space. The parallelogram equality
gives that ^ > ^2, and L2 are Hilbert spaces. In Chapter 6, we show that ?p
and Lp are not even isomorphic to a Hilbert space for p ^ 2.
Let H be a Hilbert space. Let x,y ? H. We say that x is orthogonal to
y, denoted x JL y, if (#, y) = 0. Let M C H. We say that x is orthogonal to
M, denoted x _L M, if x is orthogonal to every vector y from M.
Definition 1.31
Let F be a subspace of a Hilbert space H. The set FL — {h G H\ hi. F}
is called the orthogonal complement of F in H.
Fact 1.32
If F is a subspace of a Hilbert space H, then FL is a closed subspace of H.
Proof: Clearly, FL is a subspace. If /, ft, hn G H, [hn, /) = 0 and hn —*• h,
then (h,f) = 0. This follows from the continuity of the functional h »->•
(ft, /) discussed earlier. Hence, FL is a closed subspace.
?
Obviously, F fl FL = {0}. Therefore, every element z ? F + FL has a
unique expression in the form z — x + y with x ? F, y ? FL. We can also
see that the orthogonality gives ||z||2 = (x + y, x + y) — ||x||2 + ||y||2. It
follows that T: F 0 -F1 —> # defined by T(a:, y) = x + y is an isomorphism
of F 0 FL onto F + F1 C H.
Theorem 1.33 (Riesz)
Let F be a subspace of a Hilbert space H. If F is closed, then F + F1- = H.

18 1. Basic Concepts in Banach Spaces
Thus, T: F © F1- —» H defined by T(x,y) = x + y is an isomorphism of
F©FX onto H.
Proof: Let x E H. We claim there is a closest element xi to x in F
and (x — x{) _L F. Let yn 6 F be such that ||z — yn||2 < d2 -f -, where
rf = dist(ar, F) = inf{11a? — z\\; z E F}. We now prove that yn is a Cauchy
sequence. Consider x — yn and x — ym in the parallelogram law:
\\2x - (yn + ym)\\2 + \\ym - ym\\2 = 2\\x - yn\\2 + 2\\x - ym\\2.
We can estimate
\\ym-ym\\2 = 2||^-2/n||2 + 2||^~t/m||2-||2^-B/n+ym)||2
= 2||z-^2 + 2||z-T/m||2-4
< 2(d2 + l)+2(d2 + X)^4d2 = l + ^
where we used m E F. Therefore, {2/n}?°=i converges to some point
a?i E F. Then lla? — xA\ = lim Ila? — yn|| = d. Put x2 — x — Xi\ note that
n—>-oo
||^2||2 = d2. Assume that X2 is not orthogonal to F, so there is z E F such
that (#2, z) > 0. Then, for ? > 0 we have
\\x - (xi + ez)\\2 = \\x2 - ez\\2 = (a?2 - ^, ^2 - ?*)
= O2, ^2) - 2ff(a?2, z) + ?2(*, *)
= d2-e{2(x2,z)-s\\z\\2).
Since (#2,2) > 0, for e small enough we have 2(x2,z) — e\\z\\2 > 0 and
therefore ||ar — (a?i + ez)\\ < d, a contradiction. Thus x2 i. F.
For any x ? H we found a?i E F, #2 ? F1 such that ar = x\ + x2) so
F + F-^ff.
?
Note that for a closed subspace F of a Hilbert space H we have (F1I =
F. Indeed, write H = F © Z, where Z = F1. If ?/ E F and z E Z, then
(y, z) = 0 and thus F C Z1. If x E if, z ^ F, write # = y + z with y ? F,
z e Z. Then z / 0, and (z,x) = (z,y + z) = \\z\\2 ^ 0. Thus x ? ZL.
Therefore, F = ZL = (F1I.
Definition 1.34
Xe< H be a Hilbert space and S C H. S is called an orthonormal set if
($1 j ^2) = 0 whenever S\ ^ s2 ? S and (s, 5) — 1 /or every s ? S.
A maximal orthonormal set (in the sense of inclusion) in H is called an
orthonormal basis of H.
Theorem 1.35
Every Hilbert space has an orthonormal basis.
Proof: If {Ma} is a chain of orthonormal sets, that is, Ma C Mp or
Mp C Ma for all indices a,/?, then (jMa is an orthonormal set. Indeed,
x —
Vn +2/m

1. Basic Concepts in Banach Spaces 19
whenever x,y ? \jMa) x ^ y, there is Mp such that x,y ? Mp and
thus (x, y) — 0. By Zorn's lemma, given an orthonormal set So, there is a
maximal orthonormal set 5 containing So.
As an immediate consequence of Theorem 1.33 we obtain that if {e^}
is an orthonormal basis of H', then span({e^}) = H. Indeed, otherwise we
find a vector x from S#-nspan({eM}) , a contradiction with the maximality
of {eM}.
Theorem 1.36
Every separable infinite-dimensional Hilbert space H has an orthonormal
basis {ei}^.
Moreover, if {ei}fl1 is an orthonormal basis of H, then for every x ? H,
CO
x — y j{x', tijei.
2 = 1
The numbers (x, e2) are called Fourier coefficients, and x = XX#, ez)e2-
is called the Fourier expansion of x or the Fourier series for x.
Proof: Let {e^} be an orthonormal basis of H by Theorem 1.35.
Since ||eM — eM>|| = \/2 for {x =? ji', by the proof of Proposition 1.27
(iii), {e^} is countable. Note that {e^} cannot be finite because H
is infinite-dimensional. Choose an ordering of {e^} into a sequence
{ei,e2,...}.
We claim that for x ? H and n ? N, we have
n
dist(z,span{ei,... ,en}) = a? - y^(a?,ez-)ez-
«=i
To this end, note that for all scalars ci, C2,..., cn we have
n 2 n n
- ]P cf-e,- = (x, x) + ^ |c»- - (a:, e2-)|2 - ^ |(ar, e;)|2
i = l
i=l
i=l
II n II2 n
Hence inf \\x — ^ CieAl = (x, a;) — ^ |(ar, et-)|2, and the infimum is
c1,...)cneK\\ i—i II i—i
CO
attained when c* = (x, e2). We now show that x — Y^,(x> e0e*• Given e > 0,
i=l
find n0 such that dist(ar,span({e2}^_1)) < ? using sp^T^e;}?^) = H. For
n > no, we have
z-^(a:,ej)ej = dist(jc,span{ei,..., en}) <
< dist(x,span{ei,.. .,eno}) < e.
i=l
?

20 1. Basic Concepts in Banach Spaces
Proposition 1.37
Let {ei}fl1 be an orthonormal set in a Hilbert space H and x G H.
oo
(i) ^2 \(x,ei)\2 < \\x\\2 (The Bessel inequality).
*=i
oo
(ii) If {ei}^ is an orthonormal basis of H, then \\x\\2 = J2 \(x>ei)\2
*=i
(The Parseval equality).
(Hi) If the Parseval equality holds for every x G H, then {e;}?^ is an
orthonormal basis of H.
(iv) If span({e2-}^:1) = H, then {^i}^zl is an orthonormal basis of H.
Proof: (i): For every n G N, we have
n 2 n
x - J^(x, ei)ei | = (ar, a:) - ]T |(s, e,)|2.
2=1 2 = 1
From this, the Bessel inequality follows.
(ii): From the proof of Theorem 1.36, for every n G N we have
n
dist(z, span{ei,..., en}) = (x, x) - J^ |(ar, e2-)|2,
2 = 1
n
and dist(x,span{ei,... ,en}) —> 0 as n —> oo. Thus (#,#) — ?^ |(xt-et-)|2 —> 0
2 = 1
as n —> oo.
(iii): If {e^ is not an orthonormal basis of H, then span{e2} ^ H. By
Theorem 1.33, there is e G -ff \ {0} such that (e, e2) = 0 for every i G N.
Then 0 = ^ |(e, e2-)|2 ^ ||e||2, a contradiction.
(iv): Assume that span{e2} = H and {e2} is not an orthonormal basis.
Choose e orthogonal to all e,- and e ^ 0. Let yn G span{e2}, yn —> e.
Then (e,yn) = 0 for every n, and therefore (e,e) — lim((e,yn)) = 0, a
contradiction.
?
Theorem 1.38 (Riesz, Fischer)
Every separable infinite-dimensional Hilbert space H is linearly isometric
tol2.
Proof: Let {e,-}?^ be an orthonormal basis. For x G H, put T(x) =
(x^ei)(^zl. Parseval's equality implies that (x,e2J- G h and that T is an
isometry. It remains to prove that T maps H onto ^2- Given (c2) G <^2>
II n ll2 n
define x G ff by x = ]?c*e*- Then clearly ]T} Cj-e» = J^ |c2-|2; therefore
2 = 1
2 = 1
Ylciei is Cauchy and thus convergent in H. Next, we will show that if
oo
x — Yl cieii then Ci = (x,e2) for every i, that is, T(x) = (c2). This follows
2 = 1

1. Basic Concepts in Banach Spaces 21
from the continuity of the inner product:
oo n
QT Ciei}ejJ - ^(X^ c*e«> ei) = ^(cy) = cj for every j.
t=i n f=i
Thus, T is an onto map.
?
Similarly, we prove that every Hilbert space is isometric to some ^(r).
Note that:
(i) The space 1^2 [0,1] endowed with the inner product (/,</) = /0 fgdt
is a separable Hilbert space. Similarly, ?2 is a Hilbert space, the inner
CO
product being ((an),(&n)) = ]C anbn> Therefore, ?2 is linearly isometric
n=l
to ^2- Note that in Chapter 6 we will show that ?p is not isomorphic to Lp
forpG [l,oo),p^ 2. The space ioQ is isomorphic (Exercise 6.18) but not
isometric (Exercise 8.31) to L^.
(ii) We have shown that if {ei}^ is an orthonormal basis of a Hilbert
CO
space #, then x — Yl(x>ei)ei f°r everv # G #, and this expression is
*=i
unique, that is, if for some c2- G K we have x = ^ ct-e{, then ct- = (a:, e,) for
every i.
(iii) If {e*}^! is an orthonormal basis of H and {e^} is a permutation
of {e;}, then XXZ> e7r(*))eT* ~ x an(^ {e7r(i)} is a basis of E. Indeed, given
e > 0, find no G N such that dist(x, spanjei,..., eno}) < e. Then find mo
such that {!,..., no} C {^"A), • • •, ^r(^o)}- For m > mo, we have
1=1
x - ]P(#, e*-(o)cir@ = dist(a?, span{eT(t-),..., e,r(mo)})
< dist(x,span{ei,. ..,eno}) < e.
Exercises
1.1 Let X be a normed linear space. Prove that for any x, y G X we have
|IWI-IMI|<II*-»II-
Hint: Triangle inequality, ||z|| = \\(x — y) + y\\.
1.2 Let X be a normed linear space. Assume that for x, y G X we have
\\x + y\\ = ||x|| + ||y||. Show that then \\ax + 0y\\ = a\\x\\ + C\\y\\ for every
a,/?> 0.
Hint: Assume a > j3. Write
\\ax + py\\ = \Wx + y)-(<*-p)y\\>a\\x + y\\-(a~l3)\\y\\
= «CII*II -+" ll»ID — C« — /7>||2#|| = "Harll -+- /?|Ii/Il-

22 1. Basic Concepts in Banach Spaces
1.3 Show that span(Z) = span(L) and conv(M) = conv(M).
Hint: By the definition, span(X) is the intersection of all closed subspaces
containing L, and span(L) is one such subspace. From this, one
inclusion follows. The other inclusion follows similarly because span(L) is the
intersection of all subspaces containing L.
1.4 Show that C is a convex set in a vector space if and only if Y2 ^ixi € C
whenever x\,..., xn G C and Ai,..., An > 0 satisfy Yl^i ~ 1-
Hint: (a) Ax x1 + A2^ + A3 xs = (Ai + X2)(j^^ xx + j^^ x2) + A3 x3
and induction.
1.5 Let A and B be two convex sets in a normed space X. Show that
conv(A UB) = {Xx + A - A)y; a? G A, y G B, A G [0,1]}.
Hint: Show first that the set on the right-hand side is convex.
1.6 Let A be a set in a real Banach space X. Show that the symmetric
convex hull of A (i.e., the intersection of all symmetric convex sets containing
{n n >j
Y XiX(] Xi G A, J2 1^1 < 1 f •
2=1 2=1 J
Hint: Show first that this set is convex and symmetric.
1.7 Let A be a subset of a Banach space X. Denote by sconv (A) the set
CO
of all x G X that can be written as x — Y A2?2-, where Xi G A) A; > 0, and
2 = 1
Y^^i — 1. Show that if A is bounded, then sconv (A) C conv(A).
Let A be the set of all standard unit vectors ez- in ?2- Show that 0 G
conv(yl) and 0 ^ sconv (A).
CO
Hint: If A2- > 0 and ^ Az- = 1, approximate by a finite linear combination
2 = 1
given by Ax = Al5..., An = An, An+i = 1 - ? A*.
2 = 1
If ]T ^*e* — 0j then ^2 = 0 for every i.
1.8 Let 1 < p < q < 00. Show that ||s||/q < ||x||/p for x G ?p and ||/||Lp <
||/||Lgfor/GLjO,l].
In particular, ?p C tq, Lq[Q, 1] C Lp[0i 1], and the corresponding identity
operators have norm one.
Hint: Take x = (xi) G ^p such that \\x\\ip = 1. Then |#2| < 1, so |#;|? <
\xi\p. Consequently, ||a?||| < \\x\\* = 1. The inequality for function spaces
follows from the Holder inequality used with r = q/p.
1.9 Let / G LPo[0,1] for some p0 > 1. Show that lim ||/||Lp = ||/||Ll. If
/ G Loo[0,1], then lim \\f\\Lp = ||/||Loo.
p—»-oo
Let x G <?« for some q > 1. Show that lim |HU_ — IH^-
p—»-co p

1. Basic Concepts in Banach Spaces 23
Hint: If / is bounded, we have that fp—>f pointwise as p —> 1. By the
Lebesgue dominated convergence theorem, fQ \f\pdt —> /0 \f\dt. We may
assume that |/| < 1, then ft \f\p dt < 1, and hence J* \f\p dt < \\f\\Lp <
fo l/l dt\ consequently ||/||Lp -» ||/||Ll as p -> 1.
If / is not bounded, given ?, we first find a bounded function F G LPo
such that ||/-F||lP0 < e. Then also \\f-F\\Lp < e for every p G [l,Po] and
we can use the preceding result to find 6 > 0 such that |||/||z,i — ll/IU* | < 3?
whenever p G A,1 + 6).
Let / G Loo[0,l]. Given e > 0, set M = {t G [0,1]; \f(t)\ > \\f\\Loo -
e}. By the definition, A(M) > 0. We have ||/||Lp > {fM\f\p d\I/p >
A(MI^(||/||Loo - e). Since A(MI^ -> 1 as p -+ oo, there is p0 such that
lll/IU, - ll/IUoo I < 2e for p > po.
Now assume that x G ?q and ||«||oo = 1- Then there is a finite set M of
coordinates i such that \x{\ = 1, denote i? = \M\. Note that ^ \xi\p — K.
Fix n0 > max(M) such that ]T |a:2-|? < if/2. Because \x{\ < 1 for z ^ M,
2>n0
we have ]T |#2-|p < if/2 for every p > q] moreover, there is po > <Z such
2>n0
that |z;|p < 2,^^ whenever p>po and z ^ n0, i ? M. Consequently,
oo
K < Y, \xi\p <K + ? + ? = 2K
i=l
for all p > p0; that is, K1^ < \\x\\ip < BKI>p. Letting p -> oo, we get
lim \\x\\t =l = \\x\\lao.
p—>oo *
1.10 Show that for every p > 1, ?p is linearly isometric to a subspace of
LP[0,1].
Hint: Consider span{/n}, where fn = (n(n + 1)) p\[ * /]•
1.11 Show that c0(r) is the closure of c0o(r) in ^ (r).
1.12 Let T be a set and p G [1, oo]. Show that cq(T) and ?p(T) are Banach
spaces.
Hint: The proof of Proposition 1.11.
1.13 Show that ?P(I) is linearly isometric to ?p(J) whenever card(i) =
card(J). Here, card(i) denotes the cardinality of the set i.
Hint: If <p is a bijection from I onto J, consider the map / h-> / o (p.
1.14 Let Cn[0,l] be the space of all real-valued functions on [0,1] that
have n continuous derivatives on [0,1], with the norm
||/||= max (max{|/*(*)|; * G [0,1]}).
0<k<n

24 1. Basic Concepts in Banach Spaces
Show that Cn[0,1] is a Banach space.
Hint: If fn —» / uniformly and fn —> g uniformly, then f = g.
1.15 Let C be a normed space of all Lipschitz functions on a Banach space
X that are equal to 0 at the origin, under the norm
11/11 = »p{lzljgEJP;*,ve*}.
Show that ? is a Banach space.
Hint: Cauchy sequences are bounded.
1.16 Let V be a normed space of all bounded Lipschitz Frechet
differentiate functions (for definitions, see Chapter 8) on a Banach space X under
the norm ||/|| = sup{|/(x)|; x E X} + sup{||/,(ar)||; x E X}, where \\f'(x)\\
is the norm of f'(x) in X* (i.e., ||/'(a?)|| = sup{f(x)(h); h E Bx}). Show
that V is a Banach space.
Hint: Classical rules of differentiation.
1.17 Show that a normed space X is a Banach space if and only if J2Vn
converges whenever \\yn\\ < 2~n for every n.
Hint: Use Lemma 1.15. Note that if J2 xk is absolutely convergent, there
Nk+1
are Nn so that ? ||**|| < 2~*. Set yn = ? **, then ||yn|| < 2"n,
A:>iVn fc=ATn+l
m M
and if M > Nm. then
E 2/n - E xk
n=l Jb=l
<2~r
1.18 Let y be a closed subspace of a normed space X. Show that if Y and
X/Y are both Banach spaces, then X is a Banach space.
Note: A property V is said to be a three-space property if the following
holds: Let Y be a closed subspace of a space X. If Y and X/Y have V,
then X has 7* (see, e.g., [CaGo]).
Thus, the property of being complete is a three-space property in the
class of normed linear spaces.
Hint: If {xn} is Cauchy in X, there is x E X such that xn —> x. There are
{yn} in Y such that {xn — x — yn} —> 0. Thus {yn} is Cauchy, so yn —* y
and zn —* x + y.
1.19 Let Y, Z be subspaces of a Banach space X such that Y is isomorphic
to Z. Are X/Y and X/Z isomorphic?
Hint: No. Let X = ?2,Y = {@, x2, x3i...)}, and Z = {@,0, x3, x4,...)}.
1.20 Show that the distance d(x) of a point x = (z2-) E ^oo to Co is equal
to limsup \xi\. Thus, the norm in ?oo/co is ||?|| = limsup \x{\.
i—*oo 2—>oo
Hint: There are only finitely many i such that \x{\ > limsup \x{\ + e.

1. Basic Concepts in Banach Spaces 25
1.21 Let || • \\1: || • ||2 be two norms on a vector space X. Let Bi and B2
be the closed unit balls of (X, || • HJ and (X,\\ • ||2), respectively. Prove
that || • Ik < C|| • ||2 (that is, ||z||i < C ||ar||2 for all x E X) if and only if
¦q B2 C B\.
1.22 Let || • JI x and || • ||2 be two equivalent norms on a vector space X. Let
Bi and B2 be the closed unit balls of (X, || • 1^) and (X, || • ||2), respectively.
Show that Bi and B2 are homeomorphic.
Recall that two topological spaces K and L are called homeomorphic
if there exists a bijection <p from K onto L such that <p and <?>-1 are
continuous. Such a <p is called a homeomorphism.
Hint: Define a map <? from Bi onto B2 by <?@) = 0 and <j){x) = |J-z for
a;G5i\ {0}. Clearly, ||^(^)||2 — ||e||i> and continuity at 0 follows from the
equivalence of the norms.
1.23 Let X be the normed space obtained by taking cq with the norm
Ikllo = 532~*|a?j|. Show that X is not a Banach space.
Note that this shows that || • ||0 is not an equivalent norm on Co.
n
Hint: The sequence {A,1, ...,1,0,.. -)}nLi *s Cauchy and not convergent
since the only candidate for the limit would be A,1,...) ^ Co-
1.24 Let X, Y be normed spaces and T ? B(X, Y). Show that
||r|| = sup{||r(ar)||y; ||x||x < 1} = sup{||T(z)||y; ||ar||jr = 1>-
Hint: Clearly, both suprema are not greater than ||T||. Given e > 0, find
x e Bx such that ||T(ar)||y > vT^7||T||. Then Hx/T17!^!! < 1 and
^E&, and both vectors give \\T(y)\\Y > A - e)\\T\\.
1.25 Assume that T is a linear operator from a normed space X into a
normed space Y such that {T(xn)} is bounded for every sequence {xn} C X
satisfying \\xn\\ —* 0. Is T necessarily continuous?
Hint: Yes. Assuming the contrary, consider a sequence {a?n/\/lkn||} for
{xn} such that xn —> 0 and T(xn) -/+ 0.
1.26 Let T be a one-to-one bounded linear operator from a normed space
X into a normed space Y. Show that T is an isometry onto Y if and only
if T(BX) = BY if and only if T(SX) = SY if and only if T(B%) = JB?,
where B% is the open unit ball in X.
Hint: By homogeneity, T is an isometry onto Y if and only ifT(Sx) — Sy.
Assume that T(Sx) = 5y. If there is x ? Sx such that ||T(z)|| = C < 1,
then ||a:/C|| > 1 and ||T(z/C)|| = 1. But there must be y E Bx such that
T(y) = T(x/C), a contradiction with T being one-to-one.
1.27 Let X,Y be Banach spaces and T G #(X, Y). If there is 6 > 0 such
that ||r(ar)|| > «||a:|| for all x eX, then T(X) is closed in Y. Moreover, T
is an isomorphism from X into Y.

26 1. Basic Concepts in Banach Spaces
Hint: Let {T(xn)} C T(X) be such that T(xn) -+ y. Then {T(xn)} is
Cauchy and hence also {xn} is Cauchy. By completeness,
T(xn) —+ T(x), by the uniqueness of the limit we get y — T{x).
The inequality we assume implies that T is one-to-one, so the inverse
T-^rpQ -+ X is denned and \\T~l\\ < f
1.28 Let X be a finite-dimensional normed space, and let Y be a normed
space. Let T be a bounded linear operator from X onto Y. Assume that
there is 6 G @, |) and a finite <S-net M = {#;} in Sx such that A + <5)_1 <
l|r(a?0ll < C1 +6) for every *'• Then T_1 exists and ll^llll^!! < °(S) =
Vl-EAl+<5 1-6 /
Hint: Given x G Bx> there is X\ G M with <$i = ||# — #i||x < <$• Then b\M is
a<$i<$-net in <5iSx, so there is X2 € M such that #2 = \\x—x\ — 61X21
II n
By induction there are Xk ? M and <$& < <5* such that \\x— ]T <$fc?fc
<<5n.
Jb = l
Thus x — Yl faxk, hence by continuity ||T(a?)||y < A + 6) ]T $k and thus
ll^ll < i^f • Then also for x G Sx and a suitable Xk G M, we have
\\T{x)\\Y > \\T(xk)\\Y - \\T\\ \\x - xk\\x > ?g - *i±f.
1.29 Let M be a dense subset (not necessarily countable) of a Banach
space X. Show that for every x G X \ {0} there are Xk G M such that
z = ?a?fc and ||^|| < 5M.
Hint: Find x\ G M such that ||z - a?i|| < ^; then by induction find
Xjfc G M such that ||a?-(a?i + ... + Xk-i) - Xk\\ < ^. Then x — ^Xk and
IMI = ||(x-E1^)-(x-Ea:n)||<J& + ^ = M#.
11 v n=l y v n=l ' •'
1.30 Show that a Banach space X is separable if and only if Sx is
separable.
Hint: If {xn} is dense in Sx, consider {rkXn}k}n for some dense sequence
{nj in K. If S is countable and dense in X, consider {jffjr; # G ?}.
1.31 Let y be a closed subspace of a Banach space X. Show that if X is
separable, then Y and X/Y are separable. Show that if Y and X/Y are
separable, then X is separable.
Thus, separability is a three-space property.
Hint: If {xn} is a dense set in X/Y and {zn} is dense in Y, choosing
yn G ?n and considering {yn -\- Xk] n,k G N}, we have a dense set in X.
1.32 Find two (obviously not closed) subspaces F\ and F2 of a Banach
space X such that i*\ H F2 = {0} and both i<\ and F2 are dense in X.
Hint: X=C[0,27r], let i*\ be all polynomials on [0,27r] and F2 be all
trigonometric polynomials on [0, 2tt].

1. Basic Concepts in Banach Spaces 27
1.33 Show that the space BC(Q, 1) of bounded continuous functions on
@,1) with the sup-norm is nonseparable.
Hint: Fix {/n} in BC@,1) such that supp/n C [^, ?] and Wf^ = 1,
and then define T:4o -+ J3C@,1) by T((a„))(t) = anfn(t) for * G [^, ?).
Thus, BC(Q, 1) contains an isomorphic copy of i^. Note that this cannot
be done in C[0,1], because linear combinations Ylanfn are not necessarily
continuous at 0 or 1.
1.34 Let Ylxi ^e a series in a Banach space X, x ? X. Show that the
following are equivalent:
(i) For every e > 0, there is a finite set F C N such that
E xi
ieF'
<e
whenever F1 is a finite set in N satisfying F' D F.
(ii) If 7r is any permutation of N, then ? x*(i) — x-
If these conditions hold, we say that the series is unconditionally
convergent to #. As in the real case, a series ]T) X{ is unconditionally convergent if it
ii ii °°
is absolutely convergent: ^ ^ < ]T \\xi\\ for G C N with n < min(G).
"zEC? " i=n
Note that the unconditional convergence of a series does not in general
imply its absolute convergence; consider E ^ei in ^2-
Hint: (i) => (ii): For every e > 0, get a finite i7" such that \\J2 xi ~~ # < ?
II j./ II
whenever F' is a finite set in N such that F' D F. For some no, we get
{<!), irB), -.., 7r(no)} D F. Thus x - ? **(,-)
t=i
< e for n > no.
(ii) z=> (i): By contradiction, get by induction a sequence n^ such that
nk
Y,xi ~
t=l
< t and a sequence of finite sets M& such that rik+i >
> e.
max(Mfc) > min(Mfc) > n^ in such a way that ? X{ + ? X{ — x
"« = 1 Mfc
Indeed, since (i) fails for F = {1,..., nfc}, we find i*1' and set Mk = F'\ F.
Find a permutation n such that tt({1, ... ,njb + |Mfc|}) = {1,... , nfc}UM&;
then ? ^7rB) does not converge.
To see that ]P ie,- converges unconditionally in ^2, note that ? ie;
11 G
small if min(G) is large enough.
is
1.35 We say that Y2xi ls unconditionally Cauchy if, given e > 0, there is
< e whenever F1 is a finite set in N
a finite set F in N such that
X>2
F'
satisfying F' Pi F = 0.
Show that a series ? a?* in a Banach space X is unconditionally Cauchy
if and only if it is unconditionally convergent.

28 1. Basic Concepts in Banach Spaces
Hint: If J2xi ls unconditionally Cauchy, then it is Cauchy and thus it
converges to some x G X. Given e > 0, find a finite F\ such that
F>
< 6
for every finite F' satisfying F' fl F\ = 0. Then find n0 > max(i<\) such
that
?a?i-x <e and set F = {1,... ,n0}. If F' D F, then {i G F'; i >
*=i "
no
n0} n Fi = 0, so \\J2xi - ar
<
/ v #i #
+
E **•
2 = 1
<2e.
1.36 Show that the following are equivalent:
(i) J2xi *s unconditionally convergent.
(ii) Ylxni is convergent for every increasing sequence {rii}fl1.
(iii) J26ixi converges for every a = ±1.
Hint: (i) => (ii): Given ? > 0, there is no such that E^m < e whenever
II a II
a finite a has the property that min(cr) > no. If k, I > no, then nk^nj > no
II ' II
and thus E #ni < ?. Thus E xn{ is convergent.
(ii) => (i): If E ?* is not unconditionally Cauchy, then there is e > 0 and
a sequence F{ of finite sets such that max(Ffc) < min(Ffc+i) and E^M ^
"Fk "
e. Then enumerate the sets F& into a sequence {n^} to get that Y2xnt is
not Cauchy.
(i) => (iii): Given ? > 0, get a finite F such that E ^' ^ e wnenever
II a II
a Ci F = 0. Let no > max(F). Then, for m > n > no we have that
{n, n + 1,..., m} Pi F = 0 and therefore
IV^
<
n<i<m,e;=l
+
?
~a?j
n<z"<m,e:i=—1
<2e.
Thus, Yl€*xi ls Cauchy and hence convergent.
(iii) => (i): By contradiction, assume that Ylx* is n°t unconditionally
Cauchy. Thus, there is e > 0 and finite sets Fk such that max(Ffc) <
min(i<fc_|_i) and E^' > e' Define ei — 1 f°r * ? U^fc and Si = —1
11 Fh »
for i <? [jFk} let nfc = min(Fjb). Then E x» + E ?ixi\
'U'=nfc+'l i=nfc + l '
E *;
> 2s:, so either
E xi
i=nk + l
> e for infinitely many ?,
contradicting the convergence of E^'j or E ?ixi
"i=nk + l
ky contradicting the convergence of E ?*"#*'•
"fc+i
> e for infinitely many

1. Basic Concepts in Banach Spaces
29
1.37 Let Y2xi be an unconditionally convergent series in a Banach space
X. Show that for every e > 0 there is no such that
m
Si — ±1 and ra > n > no; in particular,
7 v ?i ¦?«
7 v SiXi
i—n
< e.
< e for every
Hint: See the previous exercises.
1.38 Let ^Xi be unconditionally convergent. Show that:
(i) Si = \ Y2 tixi\ ei — il f is a compact set in X,
(ii) S2 — \Y1 eixi\ n ? N, ?» = ±1 f is a relatively compact set in X.
Hint: (i): The space {—1,1}N is compact in the pointwise topology. We
claim that the map {?,} »-» J2e*xi ^s continuous. Indeed, from the
unconditional Cauchy condition we have that if a is a finite set with min(cr)
sufficiently large, then MT) xA\ < ?, which gives that if no is sufficiently
i?a
large, then
7 v ?*2Ti
< ? for every possible €{.
(ii): Use (i) and write 2 ^ ?;?; = ]T ?T»a?i + ]T) ?$•?;, where ^ = ?z- for
i — 1,..., n and ?[• = —?2- for i = n + 1, n + 2,
1.39 Let H be a Hilbert space. Prove the generalized parallelogram
equality: If xi,..., xn ? H, then
E |EHf=2nl>.-ii2.
e»=±l z = l i=l
Hint: Induction on n.
1.40 Let X be a Banach space whose norm || • || satisfies the parallelogram
equality. Define (x, y) by the polarization identity, and prove that (x: y) is
an inner product.
Hint: Clearly, (•,•) is continuous in both coordinates, (x,y) — (yyx) and
(—#, y) — — (#, y). Using the parallelogram equality, show that (x -\-y,z) =
[x) z) + (y, z). Then by induction, (nx, y) — n(x, y) for all n E N, and hence
also for all integers n.
Given ?, we have (?s,y) = n(?x,j/) = %m(±x,y) = ?(?s,y) =
^(x, y). By continuity, we get (ax, y) = a(x, y) for all a?R.
1.41 Find a Hilbert space H and its subspace F such that H ^ F -\- F1.
This shows that the assumption of closedness in Theorem 1.33 is crucial.
Hint: Consider the subspace F of finitely supported vectors in ?2- Then
FL = {0} because given x E H \ {0}, (x, e2) / 0 for i E supp(x).

30 1. Basic Concepts in Banach Spaces
1.42 Let H be a Hilbert space. Show that there exists an abstract set T
such that H is isometric to ^(r).
Hint: Take an orthonormal basis {e7}76r and follow the proof of
Theorem 1.38.
1.43 Show that a series Ylxi m a Hilbert space is unconditionally
convergent if and only if ]T ||^i||2 < oo.
Hint: Use Exercises 1.37 and 1.39.
1.44 Suppose {xk}<^L1 is an orthonormal sequence in ^ where xk — (xk).
Show that lim (x\) = 0 for every i G N.
k—»-oo
Hint: Use the Bessel inequality to show that (e2-, xk) —* 0 as k —> oo.
1.45 A Banach space X is said to have cotype 2 if there exists a constant
C > 0 such that for all vectors xi>..., xn ? X we have
^?||EHI4(?W!)'P
?t=±l 2 = 1 2 = 1
We say that X has type 2 if there exists a constant C > 0 such that for all
vectors a?i,..., xn G -X" we have
^E|E^i|<^(EiKH2I/2-
271
?i=±l 2 = 1 Z = I
The type/cotype q is defined similarly.
Taking for granted (see, e.g., [MiSc] or [T-J]) that the average on the
/ II n ||2\1/2
left-hand side can be replaced by the expression ( ^ ]P ^ SixAl J ,
show that if X is a Banach space isomorphic to a Hilbert space, then X is
both of type 2 and cotype 2.
A theorem of Kwapieri says that every space that is of type 2 and cotype
2 is isomorphic to a Hilbert space ([Kwa], see, e.g., [T-J]).
Hint: Square both sides of the inequality and use Exercise 1.39.
1.46 Show that ?4 is not isomorphic to a subspace of ?2.
Hint: Show that ?4 is not of cotype 2 by considering the standard unit
vectors.
1.47 Let X be a normed space, M C X and e > 0. We say that A C M
is an s-net in M if for every x G M there is y G A such that ||a? — y\\ < e.
We say that icMis e-separated in M if \\x — y\\ > e for all x ^ y G -A.
Clearly, a maximal e-separated subset of M in the sense of inclusion is an
er-net in M.

1. Basic Concepts in Banach Spaces 31
Let X be an Tt-dimensional real normed space. Show that if {xj }j_.^ is an
?-net in Bx, then N > e~n. On the other hand, there is an ?-net {xj}jL1
for Bx with N < (f+ l)n.
Hint: Fix a linear isomorphism T of X onto Rn. For a compact subset C
of X, we define the volume of C by vol(C) = A(T(C)), where A is the
Lebesgue measure on Rn. Let B(x, r) be the closed ball centered at x with
N
radius r. We have Bx C (J B(xj}e), so
;=i
AT N
vol(Bx) < vol(\J B(zj3e)) <Y^vol(B(Xj,e))
= N • vol(eBx) = AT • en • volEx).
Thus iV>6:-n.
On the other hand, if {xj}jL1 is a maximal ^-separated set in Bx, then
{^jlj^i is an ^net ^n •?*• Since B(xj,e/2) n B(xi,e/2) — 0 if z ^ j and
B(a?j, e/2) C A + ?/2)Bx, we have
iV
N-(e/2)n.vol(Bx) = vol(|J B(xhe/2))
< vol((l + e/2)B*) = A + ?/2)n volEx).
ThusiV<(ii^r.
1.48 Show that a bounded set M in cq is totally bounded if and only if for
every e > 0 there is no such that |a?n| < e for every x G M and n > no-
Formulate and prove the analogous result for lv spaces.
Hint: Every bounded subset of Rn° is totally bounded.
1.49 The Hilbert cube C is defined as {x = (&,-) G 4; Vi: \xi\ < 2~{}. Show
that the Hilbert cube is a compact set in ?2-
oo
Hint: Given e > 0, there is no such that ]T) |x2-| < ? for every x G C.
i=n0+l
Then use finite e-nets in Rn°.
1.50 Show that the set A = {a: G ^2i XX ^ + lJ^ — 1} does no^ contaAn
an element with norm equal to sup{||a:||; x G A}.
Hint: If x G A, then ||z||2 = ? s? < ?(l + 7)*? < 1 = suPa 11*11-
1.51 Let C be a convex set in a normed space X) assume Int(C) / 0
(recall that Int(C) denotes the interior of C). Show that Int(C) = C and
Int(C) = Int(C).
Hint: Use the "cone" argument. There is a point xq and an open ball Bf
such that xq + Bf C C. Note that if we take any c G C, then by the

32 1. Basic Concepts in Banach Spaces
convexity of C, the cone with the vertex c and the base xo -f Bf is a subset
of C, and all points in this cone but the vertex c are its interior points
(draw a picture!). Therefore, given c ? C, we can find points from Int(C)
arbitrarily close to c, showing that C C Int(C).
Given c ? Int(C), there is e > 0 such that c + ?.??x C C. Let d =
c-h?(c—a?o). Then d ? C, so for every v > 0 there is y ? C with ||d—y\\ < v.
If v is small (namely v < / e5 ==), c is in the cone with the base
V^ + INo-cH2
#o + Bf and vertex y, so c ? Int(C).
1.52 Let A be an open set in a normed space X. Show that conv(A) is
open.
Hint: Given {a?*}"=1 ? A and {Ai}?=1 such that A2- > 0 for every i and
J^Xi = 1, note that if Ai > 0 and 0\ is an open set containing x\ and
contained in Ai then conv(A) contains the open set AiOi+A2^2+- • -+An?n.
1.53 Is the convex hull of a closed set in R2 closed?
Hint: No, check A = {(x, ?); a: > 0} U {@,0)}.
1.54 Let K, C be subsets of a normed space X.
(i) Show that if K, C are closed, K + C need not be closed.
(ii) Show that if K is compact and C is closed, then K + C is closed. Is
conv(# U C) closed?
(iii) Show that if K is compact and C is bounded and closed, then the set
conv(i? U C) is closed.
Hint: (i): Consider K = {(x, 0); a: ? R} and C = {(a:, ^); ar > 0}.
(ii): If a?n = kn + cn —» y for fen ? K, cn ? C, then by compactness
assume kn -* k\ then also cn = xn — kn -+ (y — k) and use that C is closed.
For a negative answer to conv(/<' U C), see the previous exercise.
(iii): If xn = An?n + A - An)cn —> a? for A?n ? K, cn ? C, A ? [0,1],
find a subsequence nz- such that kni —* k and An? —*• A. If A = 1, then by
boundedness A — Ani)cni —> 0, and x = fc ? if. If A ^ 1, cni —> ^y- ? C
by closedness of C.
1.55 Let A,B be convex compact sets in a Banach space X. Show that
conv(A U 5) and A + B are compact. Generalize this statement to a finite
number of sets.
Hint: Using Exercise 1.5, show that conv(AUi?) is a continuous image of the
compact set {(a,/?); a,/?>0,a+/?= l}xAxB, so it is compact. Similarly,
A + B is the image of A x B under the continuous map (a?, y) h-» x -f- y.
1.56 Let A be a totally bounded set in a Banach space X. Show that
conv(A) is totally bounded. Therefore, conv(A) is compact.
Hint: Let J5 be a finite <5-net for A. It is straightforward to check that
conv(i?) is a <5-net for conv(A). Since conv(J5) lies in a finite-dimensional

1. Basic Concepts in Banach Spaces 33
space and is bounded, it is totally bounded, and its <$-net would produce a
2<5-net for conv(A). Note also that the closure of a totally bounded set is
totally bounded.
1.57 Let X be a Banach space and C be a compact set in X. Is it true
that conv(C) is compact?
Hint: Not in general. Consider C = { je2}u{0} in ?2, where e* are the
standard unit vectors. Clearly, C is compact. The vector B_zj) is in conv(C)
and it is not in conv(C), since any point in conv(C) is finitely supported.
1.58 Let C be a compact set in a finite-dimensional Banach space X. Show
that conv(C) is compact.
Hint: If a point x lies in the conv(E) C Rn, then x lies in the convex hull of
some subset of E that has at most n + 1 points. Indeed, assume that r > n
and x = Y^tixi is a convex combination of some r + 1 vectors x% G E. We
will show that then x is actually a convex combination of some r of these
vectors. Assume that t{ > 0 for 1 < i < r + 1. The r vectors Xi — xr+\
for 1 < i < r are linearly dependent since r > n. Thus, there are real
r+l r+l
numbers a2- not all zero such that ^ a{X{ = 0 and ^ a2- = 0. Choose m
*=1 * = 1
so that \a{/ti\ < \am/tm\ for 1 < i < r + 1 and define c2- = t{ — 9Li^m- for
1 < i < r -h 1. Then Ci > 0, ^2 C{ = ^,U = 1, x — ^ C{Xi, and cm = 0.
Having this, we can use the fact that in Rn, conv(C) is an image of the
compact set S x Cn+1 in Rn+1 x Cn+1, where S is formed by points {A J"+1
such that Xi > 0 and J^ A2- = 1, under the map ({A}, {#;}) *-> Yl^ixi-
1.59 Let {Ki} be a finite family of convex compact sets in Rn such that
every subfamily consisting of n + 1 members has a nonempty intersection.
Helly's theorem asserts that then f]Ki ^ 0. Prove it for n = 1.
Show an example for n = 2 that n + 1 is necessary.
Hint: Consider the interval between the maximum of the left endpoints of
K{ and the minimum of the right endpoints of K{. For the general case, see
for example [DGK]. Example: three lines forming a triangle.
1.60 Let X be a Banach space. Show that if A C X is totally bounded, then
there is a sequence {xn} G X such that xn —> 0 in X and A C sconv {xn}
(see Exercise 1.7).
In particular, for every compact subset A of X there exists a sequence
{xn} such that xn —* 0 and ^4 C conv{#n} (Grothendieck).
Hint: We set A\ = A, let #i be a finite 2~2-net in A\. If Ai and 5i were
defined for i < n, let An+i = (An-Bn)f]2~2nBx] note that every o,n G An
is of the form an = an+i-{-6n, where an+i G An+i, bn ? Bn. Let Bn+i be a
finite 2_2n-net in An+i. Therefore, every element a G A = A\ is of the form
n
a = 6i + a2 = 6i + b2 + a3 = ... = Y^^n + an+i. Since an+i G 2~2njBx,
l

34 1. Basic Concepts in Banach Spaces
oo
we have a — J3 6,- = ^2B262). Then it suffices to take for {xn} the
n = l
sequence that contains first all vectors from 21B\) then from 22i?2 and so
on. Since 2*'6,- G 2iBx C 2*Ai} we have \\2%\\ < 22~l and ||zn|| -> 0.
Note that we in fact proved that if xn —* 0, then the closed convex
{oo oo >l
n=l n=l J
1.61 Let X be a Banach space and G be a subspace of X that is a G$ set
in X. Show that G is closed in X.
Hint: Let S = G\G, where G denotes the closure of G. Since G is a G<j
set in X, G is Gs in G. Hence G = f] Gn, where Gn are open subsets in G.
Therefore, 5 = [J(G \Gn) and each Gn is dense in G because it contains
G. Thus, each G \ Gn is nowhere dense in G, so S is of first category
in G. We will show that 5 is an empty set. If xq G 5, consider the set
G* = {xo + x;x ? G}. Note that G* C 5. Indeed, any point xq+x, x ? G,
is in G since G is a linear set. If for some x G G we have ?o + # G G, then
by linearity of G we have xq G G, a contradiction. Therefore, G* C 5 and
G* is of first category in G. Thus, the shift G of G* is also of first category
in G. Hence, the whole G as a union of G and 5, which are both of first
category in G, is of first category in itself. This is a contradiction, since G
is a complete metric space.
1.62 Let X be a normed linear space X, Show that if X is topologically
complete in its norm topology (that is, X is homeomorphic to a complete
metric space), then X is a Banach space.
Hint: If X is topologically complete, then by the Alexandrov theorem, X is
G<$ in the completion of X, which is a Banach space. Note that this means
that a non-complete normed space cannot be homeomorphic to a Banach
space.
1.63 Let X be an infinite-dimensional Banach space. Show that there is no
translation-invariant Borel measure /ionI such that pi(U) > 0 for every
open set U and such that fi(Ui) < oo for some open U\.
Hint: Every open ball contains an infinite number of disjoint open balls of
equal radii (see Lemma 1.23).
1.64 Let X be an infinite-dimensional Banach space. Show that X admits
no countable Hamel (algebraic) basis.
Therefore, coo cannot be normed to become a Banach space.
Hint: If {e2} is a countable infinite Hamel basis of a Banach space X)
put Fn = span{ei,..., en}. Fn are closed and thus, by the Baire category
theorem, at least one Fno has a nonempty interior; that is, there is x G X
and a ball B = SBx such that x + B C Fno. Using linearity of Fno, we

1. Basic Concepts in Banach Spaces 35
have that -x + B C FnQ) so5c(x + 5) + (-a: + B) C Fno. Thus 0 is an
interior point of Fno. This would mean that Fno — X, a contradiction.
1.65 Let X be an infinite-dimensional separable Banach space and {e7}
be a Hamel basis for X. Define a norm || • || on X by |||z|| = ^ |ar7| for
x = Y2xiei- Show that ||| • ||| is indeed a norm on X. Prove that it is not
equivalent to the original norm of X.
Hint: Consider the question of separability of X in ||| • ||| using the previous
exercise.
1.66 Let H be an infinite-dimensional separable Hilbert space. Show that
H admits a norm that is not equivalent to the original norm.
Hint: If {e{}fl1 is an orthonormal basis of H) put \\x\\ — Yl^~l\xi\ f°r
x — J2xiei- Check this norm on {e;}.
1.67 Show that the linear dimension (the cardinality of a Hamel basis) of
the space fp,pG [1, oo), is the continuum c.
Hint: First, note that card(^) = c. Therefore, the linear dimension of ?p
is less than or equal to c. On the other hand, note that if A < 1, then the
vector (A, A2,...) G ?p, and these vectors form a linearly independent set.
1.68 Find a vector space X with two norms on it such that both of them
are complete norms and they are not equivalent.
Hint: Take a vector space V of linear dimension c, and let T\ and T*i be
linear bijections of V onto ?2 and ^4, respectively. Define norms on V by
IMIi = ||Ti(aO||2 and ||z||2 = ||r2(a?)||4. Then (VJHIi) is isomorphic to
?2 and (V, || • ||2) is isomorphic to ?4. Since ?2 is not isomorphic to ?4 (see
exercises above), || • ||x and || • ||2 are not equivalent.

2
Hahn-Banach and Banach Open
Mapping Theorems
A real-valued function p on a vector space X is called a positively
homogeneous sublinear functional if for all x,y ? X and a > 0 it satisfies
p(az) = ap(x) and p(# + y) < p(x) + p(y).
If, moreover, p(ax) — \oc\p(x) for all x G X and scalars a, then p is called
a seminorm on X. Note that every norm is a seminorm.
By a linear functional on a vector space V we mean a linear map from
VtoK.
Theorem 2.1 (Hahn, Banach)
Let Y be a sub space of a real linear space X, and let p be a positively
homogeneous sublinear functional on X. If f is a linear functional on Y
such that f(x) < p(x) for every x G Y, then there is a linear functional F
on X such that F = f onY and F(x) < p(x) for every x G X.
Proof: Let V be the collection of all ordered pairs (M',/'), where M'
is a subspace of X containing Y and /' is a linear functional on M' that
coincides with / on Y and satisfies /' < p on M'. V is nonempty because
it contains the pair (Y, /). We partially order V by (M', /') -<: (M", /") if
M' C M" and /"|M, = /'. If {MaJa} is a chain, then M' = {jMa and
a linear functional /' on Ml defined by f(x) = fa(x) for x G Ma satisfy
(Ma, fa) -< (Mf, f) for all a. By Zorn's lemma, V has a maximal element
(M, F). We must show that M = X.
Assume M ^ X, pick a?i El\M, and put Mx = span{M, a?i}. We
will find (Mi, Fi) G V such that (M, F) -< (Mi, Fi), a contradiction. For a

38 2. Hahn-Banach and Banach Open Mapping Theorems
fixed a G R, we define F\(x + txi) = F(x) + ta for x G M, t G R. Then F
is linear. It remains to show that we can choose a so that F\ <p.
Due to the positive homogeneity of p and F, it is enough to choose a
such that
F\(x + xi) < p(x + xi) r ., , x
F%-Xl)fpl-Xl) for every «€M. (*)
Indeed, for it > 0 we then have
F!(ar + tx1) = tF1(f + x1) <tp(f +Xl) =p(x + tx1))
and for t — —77 < 0 we have
Fi(* + tei) = F1(x-7]x1) = TjF1(^-x1)
< w(f - *i) = 2>(z - *7*i) = p(x + txi).
But (*) is equivalent to a — F\{x{) < p(x + x\) — .F(a?) and —a =
—F\(xi) < p(x — xi) — F(x) for every x G M. This in turn is equivalent to
p(x + zi) - F(x) >a> F{y) - p(y - xx)
for every x,yG M. Thus, to find a suitable a G R we need only show that
ini{p(x + x\) - F{x)\ x G M} > sup{F(y) - p(y - xi); ye M}. This is in
turn equivalent to the statement that for every x,y G M we have
p(x + xx) - F(x) > F(y) - p(y - xi).
The latter reads F(x + y) < p(x + xi) + p(y — a?i), which is true because
F(x + y) < p(x + y) = p(x + o?i + y - xx) < p(x + xx) + p(j/ - a?i).
This completes the proof of Theorem 2.1.
?
Definition 2.2
Zd (X, || • ||) 6e a normed space. By X* we denote the vector space of all
continuous linear functional on X, endowed with the canonical dual norm
defined by ||/||* = sup |/(x)|. X* is called the dual space of X.
x?Bx
The dual norm || • ||* is often denoted by || • ||. On the other hand, the
notation || • \\x* or || • ||* is sometimes used to emphasize that we work with
the dual norm. Note that X* ^ 0 since always 0 e X*.
We can see that X* = B(X, K), Since K (that is, R or C) is complete,
using Proposition 1.19 we readily obtain
Proposition 2.3
X* is a Banach space for every normed space X.
Recall that if / G X* and x eX, then \f(x)\ < \\f\\ • ||a?||.

2. Hahn-Banach and Banach Open Mapping Theorems 39
Before we pass to normed-space versions of the Hahn-Banach theorem,
we must establish some relation between the real and the complex normed
spaces.
Let X be a complex normed space. The space X is also a real normed
space. We will denote this real version of X by Xr.
On the other hand, if X is a real normed space, then X x X becomes a
complex normed space Xc when its linear structure and norm are defined
for x, y, u) v G X and a, b G R by
(x, y) + (u, v) - (x + u, y + v)
(a + ib)(x, y) = (ax - by, bx + ay)
\\(x, y)||c = sup{|| cosF)x + sin(%||; 0 < 0 < 2tt}.
The set X x {0} = {(?,0); x G X} is a closed R-linear subspace of
Xc, which is—as a real space—isometric to X under the map (z, 0) ¦-» x.
Conversely, Xc - {h + ik\ h, k G X x {0}}.
We will verify that || • ||c is actually a norm on Xc- It is clear that || • ||c
is nonnegative, satisfies the triangle inequality, and factors real constants
to their absolute value. If a is real and z = (#, y) G Xc, then
||e~mz||c = ||(cos(a):r -f sin(a)y, — sin(a)x + cos(aJ/)||c
= sup{|| cos(^)[cos(a)ar + sin(a)y]
+ sin(fl)[-sin(a)x + cos(a)y]||; 0 < 9 < 2ir}
= sup{|| cosF> + a)x + sin(^ + a)y||; 0 < 6 < 2?r}
= sup{|| cos(?y)x + sinG7)y||; 0 < 7] < 2ir} = \\z\\c.
Therefore, || • ||c is a norm on Xc- Since max{||x||, \\y\\} < \\(x,y)\\c <
\\x\\ + \\y\\, we have that the topology induced on Xc = X x X by || • ||c is
equivalent to the product topology induced on X x X by || • ||.
We will now relate duals of X and Xr. Consider the mapping R: X* —>
Xr* defined by R(f)(x) = Re(/(#)) for x G X, where Re(f(x)) is the real
part of f(x). We claim that it is a norm-preserving map from X* onto Xr*
and is linear as a map (X*)r —> Xr*.
To see this claim, note that if X is a complex Banach space and / G X*,
then sup \f(z)\ — sup |Re(/(z))|.
zeBx zeBx
Indeed, for all z we have |/(z)| > |Re(/(z))|, so one inequality is
clear. On the other hand, for z G Bx we write f(z) = eza\f(z)\ and
have f(e~i(*z) = e'iaf(z) = |/(z)|. Thus | Re(/(e"^z)) | = |/(z)| and
||c-«z|| = ||z||.
Now we show that R is onto Xr* . To g G Xr* we assign the functional
defined on X by G(x) = g(x) — ig(ix). Then G is linear over R, but also
G(ix) = g(ix) - ig(-x) = g(ix) + ig(x) = i(g(x) - ig(ix)) - iG(x).
Therefore G is linear over C, and hence G G X*. Moreover, i?(G) = g.

40 2. Hahn-Banach and Banach Open Mapping Theorems
Theorem 2.4 (Hahn, Banach)
Let Y be a subspace of a normed space X. If f G Y*; then there exists
F G X* such that F\y = f and \\F\\X* = ||/||y*-
Proof: First, assume that X is a real normed space. Define a new norm |-|
on X by |g| = ||/||y* • ||ar||, where || • || is the original norm of X. We have
1/0*01 ^ Ilk III ^or all ? G Y, so by Theorem 2.1 there is a linear functional
F on X that extends / and \F(x)\ < ||x|| = ||/||y*|kll for every x G X.
Therefore \\F\\X* = sup{|F(x)|; \\x\\ < 1} < ||/||y*. Since F extends /, we
obviously have ||.F||x* > ||/||y* as weU- Consequently, H-FHx* = ||/||y*-
Now assume that X is a complex normed space. Consider the functional
R(f) on Yr,, where R is the isometry defined above. By the first part of
this proof, we extend R(f) to afunctional g G Xr* that satisfies ||<7||xR* =
H-ft(/)llvR* = ll/l|y- Then the norm of the functional F(x) = g(x) —
ig(ix) G X* is equal to ||(/||xR* = ||/||y*-
The real part of F is g and thus Re(F|y) = Re(/); that is, R(F\y) =
R(f). Since R is a bijection of Y* onto Yr*, we get F|y = /.
?
Corollary 2.5 (Hahn, Banach)
Let X be a normed space. For every x G X \ {0}, there is f G Sx* such
that f(x) = ||ar||.
In particular, \\x\\ = max{|/(ar)|; / G -Bx*} /or ever^ xGl.
Proof: Put Y = span{z} and define / G Y* by /(te) = t||s||. Clearly,
||/||y* = 1 and f(x) = ||z||. Using Theorem 2.4, we extend / to afunctional
from X* with the same norm.
From 1/0*01 ^ 11/11 Ikll we nave SUP 1/0*01 < Ikll- On tne otner hand,
the functional constructed above shows that the supremum is attained and
equal to ||#||.
?
Definition 2.6
Let C be a convex subset of a normed space X, and let x G C. A functional
f G X* is called a supporting functional of C at x if\\f\\ — 1 and f(x) —
sup{/(y); yeC}.
By Corollary 2.5, for every x G Sx there is a supporting functional of
Bx at x.
Consider a Banach space X. If Y is a subset of X, we define its annihilator
by YL..= {/ G X*; /(y) = 0 for all y G Y}. Note that Y1 is a closed
subspace of X*. Similarly, for a subset Y of X* we define Yj_ = {x G
X; /(a:) = 0 for every f EY}} which is a closed subspace of X.
Note that if F is a subset of a Hilbert space H, then the orthogonal
complement F1 when considered a subspace of the dual H* under the
canonical duality (cf Theorem 2.21) coincides with the annihilator FL.

2. Hahn-Banach and Banach Open Mapping Theorems 41
Proposition 2.7
Let Y be a closed subspace of a Banach space X. Then (X/Y)* is isometric
to Y1, and Y* is isometric to X* /YL.
Proof: Consider the map S:YL —» (X/Y)* defined by S(x*):x i-* x*(x).
This definition is correct, since x*(xi) = x*(x2) whenever #1,0:2 G x and
x* EYL. To see that 6 maps YL onto (X/Y)*, given / G (X/Y)*, define
x* G X* by x*(x) = /(?). Then x* G Y1 and 6(x*)(x) = a:*(a?) = /(?). To
check that 6 is an isometry, write
||«(**)|| = sup |S(**)(x)|= sup |x*(a:)| = ||a:*||.
The middle equality follows since, given ||?|| < 1, there is x G x such that
||#|| < 1. On the other hand, given ||ar|| < 1, we have ||?|| < 1.
To prove the second part of this proposition, define a map a from Y* into
X* jYL by cr(y*) = {all extensions of y* on X}. It is easy to see that <r(y*)
is a coset in X* jYL and the Hahn-Banach theorem gives that ||cr(t/*)|| =
||y*||y*. It follows that a is a linear and onto map.
?
We will now establish several separation results.
Proposition 2.8
Let Y be a closed subspace of a Banach space X. If xo ? Y, then there is
f E Sx* such that f(x) — 0 for all x EY and f(xo) = dist(a?o, Y).
We then say that xq is "separated from Y using / G X*."
Proof: Let d = dist(#o>y)- Put Z = span{y, xq) and define a linear
functional / on Z by f(x + txo) = tdist(xo,Y) for x G y and tf G K.
Clearly, /|y = 0. For u = z + tao, where aiGY and ? is a scalar, we have
\\x + tx0\\ Hf + xoll ||*o-(-f)||
<- sfe-wi.
Therefore ||/|| < 1.
On the other hand, there is a sequence xn ?Y such that \\xn — xq\\ —>
dist(z0,y)- We have d = |/(a?n) - /0&o)| < ||/|| • \\xn - a?o||, so by passing
to the limit n->oowe obtain dist(ar0jy) < ||/|| • dist(aro,y).
Thus ||/|| = 1, and /(a?o) = dist(ar0, Y). Extending / on X with the same
norm, we obtain the desired functional.
?
Proposition 2.9
Let X be a Banach space. If X* is separable, then X is separable.

42 2. Hahn-Banach and Banach Open Mapping Theorems
Proof: Choose a dense subset {/n} of Sx* • For every n G N, pick xn G Sx
such that fn(xn) > \- Let Y = span{^n}. Because Y is separable (finite
rational combinations of {xn} are dense in Y), it is enough to show that
X = y. If Y ^ X, then there is / G X\ \\f\\ = 1 such that /(a?) = 0 for
every x ?Y. Let n be such that \\fn — f\\ < |. Then
|/(xn)| - |/n(Xn) - (/n(Xn) " /W)| > |/n(xn)| " |/n(xn) " /(«n)|
> l/n(^)|-||/-/n||-|M>|-i = i
a contradiction.
?
To prove separation results for sets, we need a new notion.
Definition 2.10
Let C be a set in a Banach space X. We define the Minkowski functional
ofC, /iX:X-+ R; by jic(x) = inf{A > 0; x G AC}.
Obviously, fJ>c(x) > 0 for all x G X, but in general we can have lie — oo
on large subsets of X.
Lemma 2.11
Let C be a convex set in a Banach space X. If C contains the origin as
an interior point, then its Minkowski functional \xc is & finite positively
homogeneous sublinear functional.
Moreover, {x; fic{x) < 1} C C C {x; fJ.c{x) < 1}-
Proof: Let Bs = {x\ \\x\\ < 6} C C for some 6 > 0. Since 0 G C, the
point 0 is in AC for every A > 0 and thus /ic@) = 0. Given x G X \ {0},
we get <5|jf|y eB6 CC,so x? Mc. Thus pc(x) < ^ < oo.
Given a, A > 0, clearly x G AC if and only if ax G AaC. Therefore
/ic(az) — anci^x), and thus fie is positively homogeneous. We claim that
He is subadditive; that is, fic(% + y) < ^c(#) -f /J-c(y)- Fix any s > fic(x)
and ? > fJ>c(y)- We have that there is so < 5 such that x G soC. Note
that sqC C sC. Indeed, 0 G sC, and if c G C, then by the convexity of C,
s0c = ^(sc) + A - ^H G sC. We see that x e sC and similarly y G tC.
Then x -f y G sC -f tfC, and thus by the convexity,
Therefore Hc(% + y) < t + s, so by the choice of s and t we have /ic(# + y) <
//c(^) + //c(y)-
If /ic(«) < 1, then x G AC for some A G @,1), so \x G C. Since 0 G C
and C is convex, x = A(f) + A - AH G C. If x G C, then /iC(x) < 1 by
the definition.
?

2. Hahn-Banach and Banach Open Mapping Theorems 43
Theorem 2.12 (Hahn, Banach)
Let C be a closed convex set in a Banach space X. If xq ? C then there is
f E X* such that Re(/(ar0)) > sup{Re(/(x)); x E C}.
PROOF: First, let X be a real space. We may assume without loss of
generality that 0 E C; otherwise we consider (C — x) and xo — x for some
x E C. Let <5 = dist(xo,C). Then 8 is positive since C is closed. Set
D = {x E X; dist(a?,C) < 6/2}. Since 0 E C, we have f?x C D, so
D contains 0 as an interior point. D is also closed, convex, and xq ? D.
Let /i be the Minkowski functional of D. Since D is closed and xq (fc D, we
have fi(xo) > 1 (Exercise 2.15).
Define a linear functional on span{#o} by f(Xxo) — Xfi(xo). Then, on
span{#0}, we have f(Xxo) < fi(Xxo). Indeed, for A > 0 it is clear from the
definition of /; whereas for A < 0 we have f(Xxo) = Xfi(xo) < 0 while
fjt(Xxo) > 0.
Extend / onto X by Theorem 2.1 and denote this extension by / again.
Then f(x) < fi(x) for every x E X.
If x E D, then fi(x) < 1 and thus f(x) < ji{x) < 1 on C C D. Since
D contains a neighborhood of the origin, we have that / is bounded on a
neighborhood of zero, so / E X*. Since fi(xo) > 1 and f(xo) = /i(#o), we
get f(x0) > 1, hence f(x0) > sup{/(x); x E C}.
If X is a complex space, we construct g from X^ as in the real case and
then define f(x) = g(x) — ig(ix).
?
For simplicity, we will state the following result only for the real case.
Corollary 2.13
Let X be a real Banach space.
(i) Let C be an open convex set in X. If xq ? C, then there is f E X* and
A E R such that f(xo) = A and f(x) < X for all x E C.
(ii) Let A, B be disjoint convex sets in X. If A is open, then there is f E X*
and A E R such that f(a) < X for all a E A and f(b) > X for all b E B.
Proof: (i): We pick some y E C and consider D = C — y, yo = #o — y.
Then define fir? and / on span{yo} as in the proof of Theorem 2.12. Let /
denote the extended functional as well. We have /(yo) = A*?>(yo) > 1 since
yo ^ D. Also f(x) < 1 for x E D because D is open (Exercise 2.15), and
the statement now follows with A = 1 -f /(y).
(ii): Applying (i) to the open convex set C = A — B and Xq = 0, we
obtain / such that f(x) < /@) = 0 for x E A - B. Thus f{a) < f(b) for
every a E A, 6 E B. Set A = inf(/). Clearly, / > A on B and / < A on A.
JB
If a E A is such that f(a) — A, then from the openness of A) f(a + h) > X
for some a-f/iGi,a contradiction. Therefore / < A on A.
?

44 2. Hahn-Banach and Banach Open Mapping Theorems
Note that if A > 0 in the preceding results (for instance if 0 G C, resp.
0 G A) j by scaling we may assume that A = 1.
Duals of Classical Spaces
In Propositions 2.14 through 2.19, we assume the scalar field to be R.
Proposition 2.14 (Riesz)
Cq = ?i in the sense that for every f G cj there is a unique (ai) G l\ such
that f(x) = Y^,aixi for aH x — (xi) G Co, and the map f >-» (az) is a linear
isometry of c^ onto l\.
i
Proof: Given / G cj, define az- = /(e2), where e* = @,..., 0,1,0,...) are
the standard unit vectors in cq. For n G N, we set
xn = (sign (ai),..., sign (an),0,...) G c0.
Then Halloo = 1 and f(xn) = ? |a,-| < ||/|| • U^Uoo = ||/||. Therefore
2 = 1
OO
J2 \ai\ ^ 11/I I < °°j ^na^ ^S5 the maP / H~> (/(e0) *s a continuous map into
»=i
l\. It is obviously linear.
OO
On the other hand, if J2 |at-| < oo, then J2 \°>ixi\ < °° f°r every x —
* = 1
(x{) G c0. Indeed, we have Y^\aixi\ < sup|x2-| • ?|at-| = ||(aOI|i||(^)lloo-
Consider the linear functional ft defined on cq by ft(#) = ^a2?j. Then,
from the above estimate, we have ||ft|| < ||(a«)||i and also ft(e2) = a;,
so ft G Cq and the map / j—> (/(ez)) is thus onto. We also obtain that
11/|| = ||(/(a,-)) ||i, and hence the considered map is an isometry onto l\.
?
Proposition 2.15 (Riesz)
?* = l^ in the sense that for every f G t\ there is a unique (a2) G ^oo such
that f(x) = Ylaixi for aM x = (xo) ? ^i; and the map f *-» (a2) is a linear
isometry of l\ onto t^.
Proof: Given / G ^i, put a2- = /(e2) for i G N, where e; are the standard
unit vectors in t\. Then |a;| < ||/||, so ||(ai)||oo < ||/||. Conversely, for
(ai) G ?00, consider the functional ft defined on l\ by h(x) = J2aixi-
Again, \h(x)\ < ||(ai)||oo||«||i, and hence ft G t\ and ||ft|| < ||(a;)lloo-
Similarly as in the proof of Proposition 2.14 we conclude that the map
is a linear isometry onto t^.
?
Proposition 2.16 (Riesz)
Let p, q G A, oo) be such that - -f - = 1. Then ?* — ?q in the sense that for

2. Hahn-Banach and Banach Open Mapping Theorems 45
every f ? ?* there exists a unique element (a2) G ?q such that f(x) = ^ a{X{
for all x ~ (x{) G ?p, and the map f i—> (a2) is a linear isometry of ?* onto
Proof: For / G tp, put az- = /(ez-). Considering
xn - (KI9 sign (ai),..., lan^ sign (an), 0,...),
we see that
X>r = f(*n) < \\f\\-n = \\f\\(J2(\ai\9~1)p)' = ii/ii-(E w?) *•
2 = 1 2 = 1 2 = 1
< CO.
This reads (? |at-|«) * < ||/||. Hence ||(a2-)||? <
\=i y
If (a>i) G ^ and (a?i) G <?p, then the series Ylx%ai ^s convergent because
^|^a2-| < ||(^0llpll(a*)llg ^y ^ne Holder inequality. Therefore, the
functional h defined on ?p by h(x) = J2
well defined and \\h\\ < \\{o-i)\\q.
The rest of the proof is analogous to those above.
?
Similarly, we show that for a set T and p G [1, co) we have co(T)* = ^i(r)
and ^(T)* = tq(T), where I + \ = 1.
Proposition 2.17 (Riesz)
Letp,q G (l,oo) be such that ^ + \ — 1. TAen LP[0,1]* = Lg[0,l] m the
sense that for every F G L* there is a unique f G Lq such that F(g) =
L gf dx for all g G Lp, and the map F \-± f is a linear isometry of L*
onto Lq.
Proof: Let F G L*. For t G [0,1], let ut = X[o,t) t>e the characteristic
function of [0, i). Define a(t) = F(ut). We claim that a is absolutely continuous.
Indeed, if [t,-,^], i — 1,..., n, is a collection of non-overlapping intervals
(that is, their interiors are pairwise disjoint), put a = sign (a(^-) — a(r;))
and estimate:
n n n
J2 \<*(ti) - <*(Ti)\ = ]T ?»¦(<*(*.•) ~ a(r,-)) = F\J2ei(uti - Urj)
2 = 1 2 = 1 2 = 1
n
< ii^iU;-|I>fa<-'0|r
111=1 L"
*^0 2 = 1
2 = 1 ^ 2 = 1
Therefore, a is an absolutely continuous function on [0,1]. By the
Lebesgue fundamental theorem of calculus, we have a(t) — a@) = J0 a' dx

46 2. Hahn-Banach and Banach Open Mapping Theorems
for every t E [0,1]. Setting f — a' and using a@) = F(uq) — 0, we get
F(ut) = a(t)= / fdx- I utfdx.
Jo Jo
Since F is linear, we also have F(gn) — JQ gnf' dx for all step functions
n
k = l n
Let g be a bounded measurable function on [0,1]. Then there is a
sequence of step functions gn such that gn —> g a.e. and {gn} is uniformly
bounded. By the Lebesgue dominated convergence theorem, we get
lim (F(gn)) = lim / gnfdx= / lim (gn)f dx = / gf dx.
^°° n-*°°Jo Jo n^°° Jo
On the other hand, since gn —> g a.e. and gn are uniformly bounded, the
same theorem implies \\gn — g\\Lp —» 0 as n —> oo. By the continuity of F
on Lp, we thus have F(g) = lim (F(gn)) =: L gf dx.
n—+co
Hence, F(g) = JQ gf dx for every bounded measurable function g on
[0,1].
We will show that f ? Lq and ||/||g < ||.F||. Consider a family of functions
gn defined by
„ M-J l/WI9 sign (/(*)) if\f(x)\<n,
9n( ' -\ 0 if |/(ar)| > n.
The functions gn are bounded and measurable. Thus we have F(gn) —
Jo 9nf dx. Note also that |F(ffn)| < ||F|| \\gn\\p. On the other hand,
/ \gn\?dx = / \gn\^dx = I \gn{t)\ \g„(t)\^ dx
Jo Jo Jo
f \gn\\f\dx= / gnfdx = F{gn) = \F(gn)\.
Jo Jo
Hence ft \gn? dx < \\F\\ • ||,B||P = ||F|| (ft \gnp dx)', so (ft \gn? dx) * <
\\n
Since / is integrable, we have \gn\ —> \f\q l a.e. By Fatou's lemma, the
last inequality implies that
<
(? |/|« &)* = (jf1 M^cfe)* = (jf1 |ff„P&)* <
11*11
This shows that f ? Lq. Finally, let g ? Lp. There exists a sequence
{gn} of bounded measurable functions that converges to g in Lp. Then
F(gn) —> i^(^), and by Holder's inequality we have fQ gnf dx —» J0 gf dx.
We have shown that F(gn) = J0 #n/ c?z for bounded measurable functions,
so .F(<7) = J0 <7/g?x as claimed.

2. Hahn-Banach and Banach Open Mapping Theorems 47
On the other hand, given a function / E Lq, we can define a linear
functional on Lp by F(g) = f0 gf dx. It follows from the Holder inequality
that F is continuous and ||F|| < ||/||?.
?
Using similar methods, we obtain an analogous result for the space L\.
Proposition 2.18 (Riesz)
Lx[0,1]* = ioo[0,1] in the sense that for every F E L\ there exists a unique
f E Loo such that F(g) = JQ gf dx for all g E L\1 and the map F »—> / is
a linear isometry of L\ onto L^.
Proposition 2.19 (Riesz)
For every F E C[0,1]*; there exists a function f on [0,1] with bounded
variation such that F(g) — fQgdf (Stieltjes integral) for all g E C[0,1]
l l
and \\F\\ = \//; where \f f denotes the variation of f on [0,1].
o o
On the other hand, if f is a function of bounded variation on [0,1], then
F(g) = JQ g df is a continuous linear functional on C[0,1].
Proof: Consider the space <^oo[0j 1] of bounded functions on [0,1] with the
supremum norm denoted by || • H^. If F E C[0,1]*, we have that 1^(^I <
||F|| • l^lloo for every g E C[0,1]. Since C[0,1] is a subspace of 4o[0,1], by
the Hahn-Banach theorem we can extend F to a functional F on ^oo[0,1]
such that \F(g)\ < \\F\\ • ||#||oo- We will represent F similarly to the Lp
setting above. For t E [0,1], let ut = X[o,t), the characteristic function of
[0,t). Put f(t) = F(ut) for t E [0,1] (note that F is not denned on ut
because ut is not continuous). We will prove that / has finite variation on
[0,1]. To this end, consider to — 0 < t\ < ... < tn-i < tn = 1 and put
Si - sign (/(*,-) - /(*j_i)). We have
n n n
J2\f(ti)-f(ti-i)\ = E?<wo-/(<••-!)) = -p(Ee'-K-"*«-»))
i—l i=l i=l
n
"*=? llo°
Hence / has finite variation on [0,1], which is bounded by ||F||.
n
For g E C[0,1] and gn — J2 g{^)(uh. — Uk-i_), we have
! = 1 J°
Therefore lim (F(gn)) = lim ? *(?)(/(?) - /(*=*)) = ftgdf. Since
72 —* OO 71 "-+• OO i -i
F € 4o[0,1]* and $„ -> y in || • ||TC, we have lim(F(flr„)) = F(g), so F(g) =

48 2. Hahn-Banach and Banach Open Mapping Theorems
fQ gdf. However, for g G C[0, 1] we have F(g) — F(g), and hence F(g) =
Io9(t)df(t).
1
We have already shown that \J f < \\F\\. On the other hand, from the
o
theory of the Riemann-Stieltjes integral we have that, given a function '/
of bounded variation, F:g \—> f0 gdf is a linear map and fQ g(t)df(t) <
l l
Hsiloo V/- Therefore F is continuous and ||.F|| < V/-
o o
D
In general, if K is a compact set, the space C(K)* can be identified with
the space of all regular Borel measures on K of finite variation. Every such
measure fi defines a functional F^(f) — fK f dfi, and the correspondence
[i i-» Fp is a linear isometry ([Rud2]).
Let k G K. We define the corresponding Dirac functional (or Dirac
measure) by 6k(f) = f{k) for every / G C(K). Observe that 6k is a continuous
linear functional of norm one. Indeed, on one hand, \\6k\\ = sup Fk{f)) =
sup (/(&)) < 1. By considering the constant function / = 1, we obtain
11**11 = 1-
Proposition 2.20
The space C[0,1]* is not separable.
Proof: Consider the Dirac measures 6t for t G [0,1]. We claim that if
h ± *2j then \\6tl - 6t2\\ = 2. Indeed, \\6tl - 6t2\\ < \\6tl\\ + \\6t2\\ = 2. On
the other hand, choose /o G C[0,1] such that /o(*i) = 1, /o(^) = ~1, and
H/olloo = 1- Then ||«tl - «ta|| > |/0(*i) - fo{t2)\ = 2. Similarly to the case
of ^oo5 we find that C[0,1]* is not separable.
?
Recall that the inner product on a complex Hilbert space ?2, respectively
L2, is defined by ({xi)i(yi)) = J2xiVi> respectively (g,f) = /0 gfdx.
This motivates the following identification of the dual space in the case
of complex scalars. Recall that a map $ is called conjugate linear if
$(ax + y) = a$(x) + $(y) for all vectors x, y and scalars a.
Theorem 2.21 (Riesz)
Let H be a Hilbert space. For every f ? H*, there is a unique a G H such
that f(x) = (x,a) for all x G H. The map f *-+ a is a conjugate-linear
isometry of H* onto H.
Proof: The uniqueness of such a is clear. Indeed, if f(x) = (x,ai) —
(x,a2), then using x = a\ — a2 we get {a\ — a2,ai) = (ai — a2,a2). Thus
(ai — a2, ai — a2) = 0, so a\ — a2.

2. Hahn-Banach and Banach Open Mapping Theorems 49
By the Cauchy-Schwarz inequality,
||/||= sup 1/001= sup \(x,a)\< sup (IHI-IHD^IHI.
IWI<i IMI<i ll*ll<i
On the other hand, ||/|| = sup{|/(z)|; ||a;|| < 1} > (a/||a||, a) = ||a||. Hence
11/11 = INI-
To obtain the representation of 0 ^ / G H*, consider N = Ker(/). It is
a proper closed subspace of H. Choose zq G N1 and assume without loss
of generality that f(zo) = 1.
We claim that H — ^©spanjzo}. Indeed, given h G H, it suffices to find
a scalar a such that h — azo G N] that is, f(h — azo) = 0. This is satisfied
for a — f(h).
We now show that f(x) — (x, ,,^9,[2) for every x G H. Given x = y-fo^o,
where y G N and a is a scalar, we have
f(x) = af(z0) -a- a(z0,2r0)/||zo||2
= (y,zo)/\\zo\\2 + (azo,z0)/\\zo\\2 = fc'jTnp)-
?
Banach Open Mapping Theorem
Definition 2.22
Let (p be a map from a topological space X into a topological space Y. We
say that (p is an open map if it maps open sets in X onto open sets in Y.
Let T be a linear operator from a normed space X into a normed space
Y. Observe that if T is an open map, then T is necessarily onto. Indeed,
by Exercise 2.29, 6By C T(Bx) for some 6 > 0, and hence, by linearity,
Y C T{X). We will now establish the converse for bounded linear operators.
By Bx(r) we denote the open ball centered at the origin of a Banach
space X with radius r.
Lemma 2.23 (Banach)
Let X be a Banach space, Y a normed space, and T G B(X,Y). If r) s > 0
satisfy B${s) C T(B$(r)), then B$(s) C T(B$(r)).
Proof: By considering ~T if necessary, we may assume that r = s = 1.
Denote B% = B%A) and B$ = ?y A). Let z E B$ be given. Choose 6 > 0
such that ||z||y < 1 - 6 < 1 and put y = A - 8)~1z. Note that \\y\\Y < 1.
We will show that y G A - 6)-lT(B%), which implies that z G T(Bg).
We start with yo = 0 and inductively find a sequence yn EY such that
\\y - yn\\y < 8n and (yn - yn-i) G T(<5n_15?). Indeed, having chosen
yo,yi,...,!fe-i € Y, we have (y - yn-i) G S"-1^ C T^^), and

50 2. Hahn-Banach and Banach Open Mapping Theorems
hence there is w G T(8n~lB%) such that ||iu - (y - y„_i)||y < Sn. Setting
yn — yn_! + w we complete the construction.
Next, we find a sequence {xn}^L1 C X such that ||#n||x < <$n_1 and
T(xn) — yn—yn-i for n G N. Since the series J^ #; is absolutely convergent,
OO OO CX)
we have x = ^ #n. Then ||^||x < 5Z Ik^lU < S ^n_1 — i~j> and by
n=l n=l n=l
the continuity and linearity of T,
AT N
T(x) = Jim VT(zn) = Jim V(yn - yn_i) = Jim (yjy) = y.
n = l n=l
D
Note that T(J3^(r)) = T(Bx{r)), so the conclusion of this lemma is true
if we assume for instance 8By C T(Bx)-
Theorem 2.24 (Banach open mapping principle)
Let X,Y be Banach spaces and T G B(X,Y). IfT is onto Y, then T is an
open mapping.
Proof: Put G = T(B<x). Since T is linear, it suffices to prove that G
contains a neighborhood of the origin. Note that we have T(B^(r)) — vG
and vG — rG for every r > 0. Therefore T(B^-(r)) = rG for every r > 0.
OO
This implies that Y — T(X) — (J nG. By the Baire category theorem,
_ n=1 _
there is n G N such that nG contains an interior point, so there is x$ G G
and 6 > 0 such that (aro + By (8)) C nG. Since nG is symmetric, we have
(—xq -f By (<$)) C nG. If ? G 5y (<5), then from the convexity of nG we
have x = §(s0 + a?) + |(-*o + x) e n^- Therefore B°F) C T(J5^(n)) and
consequently B?(?) C ±T(B%(n)) = T(Bg). By Lemma 2.23, we have
?y (?) C T(B°) as claimed.
D
It follows from the proof that if T: X —> Y is onto, then there is 6 > 0
such that <55y C T(BX)-
Note that even if T G 5(X, Y) is open, it does not imply that T(M) is
closed in Y whenever M is closed in X (Exercise 7.9).
Corollary 2.25
Let X,Y be Banach spaces, and let T G B(X)Y) be onto Y.
[%) IfT is one-to-one, then T-1 is a bounded linear operator.
(ii) There is a constant M > 0 such that for every y G Y there is x G
T-^y) satisfying \\x\\x < M\\y\\Y-
(Hi) Y is isomorphic to X/Ker(T).
Proof: (i): If O is open in X, then (T'1)-1^) = T@) is open in Y,
showing that T is continuous.

2. Hahn-Banach and Banach Open Mapping Theorems 51
(ii): By the open mapping theorem, there is 8 > 0 such that 8 By C
T(Bx). Therefore, for every y ?Y, \\y\\y = <5, there is x G Bx such that
T(x) = y. Thus it is enough to put M - 1/8.
(iii): Define a linear map f from X/Ker(T) onto Y by f(x) = T(x).
Then T is one-to-one and onto Y. Let xn —> 0. Then there is arn G ?n such
that ||#n||x < ||?n|| + l/n and therefore xn —> 0. Since T is continuous,
we have T(xn) —> 0 and thus T(?n) —» 0. Hence T is continuous and
one-to-one, so by (i) it is an isomorphism of X/ Ker(T) onto Y.
D
Theorem 2.26 (Banach closed graph theorem)
Let X, Y be Banach spaces, and let T be a linear operator from X into
Y. T is a bounded operator if and only if its graph {(a^T^a?)); x G X} is
closed in 10 7.
Recall that the norm on X0Y is defined by ||(ar,t/)|| = \\x\\x + \\y\\Y- In
particular, (xn,yn) —> (x, y) if and only if xn —» x and yn —» y. Denote by
G the graph of T and note that it is a subspace ofl©7.
PROOF: If T is continuous and (xn,T(xn)) —> (zo,yo)> then yo = r(a?o).
Indeed, we have xn —> #o and T(xn) —> yo, while the continuity of T implies
that T(xn) —> T(#o). This means that (a?0, yo) = (a?o, T(a?o)) is in the graph
of T, showing that G is closed.
If the graph of T is closed in X 0 Y, then G as a closed subspace of
X 0 Y is a Banach space in the norm induced from X 0 Y. Consider the
map p: G —> X defined by p(a?,T(a;)) = a?. By the definition of the norm
in X 0 Y, we see that p is continuous, maps G onto X, and is one-to-one.
By Corollary 2.25, p'^n (a?,T(ar)) is a continuous map from X onto G.
Since also ^107-^7, g(z,y) = y, is continuous, and T = gop, T
must be continuous.
?
Definition 2.27
Zetf X,Y be Banach spaces and T G B(X}Y). We define the dual (also
called adjoint) operator T* G #(Y*,X*) /or f e Y* by T*(f):x >->
f{T{x)).
It is easy to observe that x i-» f(T(x)) is a linear map. If ||#|| < 1,
then \T*(f)(x)\ = \f{T(x))\ < \\f\\\\T\\. Thus T*(/) is also bounded, so
T*(f) G Y* and T* is well defined. Also, the map / h-> T*(/) is linear and
the above estimate shows that ||T*(/)|| < ||T|| ||/||. Consequently, T* is a
bounded linear operator from X* into Y*.
Proposition 2.28
Let X,Y be Banach spaces. IfT G B(X,Y) then \\T*\\ = \\T\\.

52 2. Hahn-Banach and Banach Open Mapping Theorems
Proof: We have
||r|| - sup \\T*(f)\\x* = sup { sup |T*(/)(z)|}
f€BY* feBY* x?Bx
= sup {sup |/(T(x))|}= sup{ sup |/(T(x))|}
f?BY* xeBx x?Bx feBY*
= sup {||r(*)||y} = ||T||.
xeBx
D
Let X, y, Z be Banach spaces, and let T G B(X, Y), S G #(y, Z). Then
(ST)* = T*S*. Indeed, consider f e Z*. Then for every x G X we get
(ST)*(/)(x) = f(ST(x)) = (S*f)(T(x)) = (T* <>*(/))(*), so ET)*(/) -
(T*S*)(/).
Exercises
2.1 Let C be a convex symmetric set in a Banach space X. Assume that
a linear functional / on X is continuous at 0 when restricted to C. Show
that the restriction of / to C is uniformly continuous.
Hint: Given e > 0, we look for a neighborhood U of the origin in X such
that x, y G C and x — y ?U imply |/(ar — y)| < ?. We have |(x — y) ? C,
so by homogeneity of / we must find U such that |/(w)| < e/2 for point
w G C fl G. Such U exists by the continuity of /L at 0.
2.2 Show that if X is a finite-dimensional Banach space, then every linear
functional / on X is continuous on X.
Hint: If {ei,... en} is a basis of X, consider the problem in the norm
||z|| = Yl \xiI f°r x — J2 xieii wnicn is equivalent to the original norm of X
by Proposition 1.22.
2.3 Show that if X is an infinite-dimensional Banach space, then X admits
a discontinuous linear functional.
Hint: Let {e7} be a Hamel basis formed by vectors of norm 1. Define a linear
functional / on {e7} so that the set {/(e7)} is unbounded, and extend /
on X linearly. Then / is not bounded on the unit ball.
2.4 Let / be a linear functional on a Banach space X. Show that / is
continuous if and only if /_1@) is closed.
Show that if / is not continuous, then /_1@) is dense in X.
Hint: Assume that / is not identically 0. Since /_1(R\ {0}) ^ 0 is open,
there is some ball B = xq + SBx such that f\B ^ 0. Assume f(xo) > 0,
then also f\B > 0 (connect xq with points of J3, / ^ 0 on the connecting

2. Hahn-Banach and Banach Open Mapping Theorems 53
segments and / is continuous). Then f\B > — jf(x0), so by symmetry of
Bx we get \f(x)\ < j/(#o) for x G Bx and / is continuous.
If/_1@) is not dense, there is a neighborhood B of some point such that
/ ^ 0 on B. Since B is convex and / is linear, / cannot be both positive
and negative on B. Assume / > 0 on B and then proceed as above.
2.5 Show that if / ^ 0 is a bounded linear functional on a normed space
X, then the codimension of /~1@) in X is 1.
Hint: The proof of Theorem 2.21.
2.6 Recall that by a hyperplane of a Banach space X we mean any subspace
Y of codimension 1 (that is, dim(X/Y) = 1).
Let Y be a subspace of a Banach space X. Show that Y is a closed
hyperplane if and only if there is / G X* such that Y = /-1@).
Hint: One direction: Exercise 2.5. Given a closed hyperplane Y, take e ? Y,
and use Proposition 2.8 to find /. Then Y C /_1@), and since codim(Y) =
1, the equality follows.
2.7 Let X be a Banach space. Show that all closed hyperplanes of X are
mutually isomorphic.
By induction, we get that given k G N, all closed subspaces of X of
codimension k are isomorphic.
Hint: Let Nf = f~l@) and Ng = g'1^). Assume Nf ? Ng. Then TV =
Nj CiNg is 1-codimensional in Ng, so Ng = iV0span{^} (algebraic sum).
Since Nf ^ Ng, we have xg ^ Nf and there is Xf G Nf such that X =
N 0 span{a^} 0 span{#/}. Assume that f(xg) = 1 = g(xf) and define
T(x) = x+ (f(x) -g(x))xf + (g(x) - f{x))xg. Then T is a bounded linear
operator on X.
For y G AT, a, /? G K, we have T(y + axf + f3xg) — y + /?a?/ + a^, so
in particular T^ is one-to-one and onto Ng, hence an isomorphism by
Corollary 2.25.
2.8 Find a discontinuous linear map T from some Banach space X into X
such that Ker(T) is closed.
Hint: Let X — cq and T(x) = (/(#), #i, a?23 • • •) for x = (#;), where / is a
discontinuous linear functional on X.
2.9 If X is an infinite-dimensional Banach space, show that there are
convex sets C\ and Ci such that C\ U C2 = X, Ci Pi C2 = 0, and both Ci and
C2 are dense in X.
Hint: Take a discontinuous functional / on X (Exercise 2.3), define C\ —
{x\ f(x) > 0} and C2 — {%] f(x) < 0}, and use Exercise 2.4.
2.10 Let X be a Banach space, / G Sx* • Show that for every x G X we
havedist^,/^)) = \f(x)\.

54 2. Hahn-Banach and Banach Open Mapping Theorems
Hint: IfyG/-1^), then \\x-y\\ > \f(x-y)\ = \f(x)\. On the other hand,
take y e Sx such that \f(y)\ > 1 - e. Then z-x- j^y G /_1@) and
dist(x, /-Ho)) < Ik - ^11 < ^^^
2.11 Let {#;}"_! be a linearly independent set of vectors in a Banach space
X and {oti^-i be a finite set of real numbers. Show that there is / G X*
such that f(xi) = ct{ for i = 1,..., n.
Hint: Define a linear functional / on span{zz-} by f(xi) — a.{ for i = 1,..., n
and use the Hahn-Banach theorem.
2.12 Let TV be a maximal ^-separated set in the unit sphere of a Banach
space X. Show that A — e)Bx C conv(A^).
Hint: Otherwise, by the separation theorem, we find x G X and / G Sx*
with ||z|| < 1 — e and f(x) > sup (/) > sup(/). For 6 > 0, choose y E Sx
convfiV) JV
such that /(y) > 1 — 6. By the maximality of N, there exists z € N with
e > ||y - 4 > f(y) ' f(*). Thus suPiV(/) > /(z) > f(y) -6>l-6-e.
This holds for any 6 > 0, so we have 1 — e < s\ipN(f) < f(x) < \\x\\ < 1 — e,
a contradiction.
2.13 Let (X, || • ||) be a Banach space. Show that fiBx{x) = ||a?||.
Hint: Use continuity of the norm.
2.14 Let A, B be convex sets in a Banach space X. Show that if t4 C 5,
then /ijg < /i^. Show that plca{x) — ^pla(x) for c > 0.
Hint: Follows from the definition.
2.15 Let C be a convex subset of a real Banach space X that contains a
neighborhood of 0 (then \iq is a positive homogeneous sublinear functional
on X). Prove the following:
(i) If C is also open, then C = {x] nc(x) < I}. If C is also closed, then
C={x- /xc(x)<l}.
(ii) There is c > 0 such that fic(x) < CIMI-
(iii) If C is moreover symmetric, then fie is a seminorm, that is, it is a
homogeneous sublinear functional.
(iv) If C is moreover symmetric and bounded, then fie is a norm that
is equivalent to || • \\x. In particular, it is complete, that is, (X,fic) is a
Banach space.
Note that the symmetry condition is good only for the real case. In a
complex normed space X) we have to replace it by C being balanced] that
is, Xx eC for all x G C and |A| = 1.
Hint: (i): Assume that C is open. If x G C, then also bx G C for some small
6 > 1, hence x G jC and fJ>c{x) < 1. Assume that C is closed. If fJ>c(x) = 1,
then there are An > 1 such that x G AnC and An —* 1. Then ^ar —* x, and
by convexity and closedness of C, x = lim(-^-x + 1~An0) G C.

2. Hahn-Banach and Banach Open Mapping Theorems 55
(ii): Find c > 0 such that \Bx C C, then use previous exercises.
(iii): Observing that \xc{—x) = Hc(x) and positive homogeneity are
enough to prove fic(^) = |A|//c(#) for all A E R, x E X.
(iv): From (iii) we already have the homogeneity and the triangle
inequality. We must show that fi.c(%) = 0 implies x = 0 (the other direction
is obvious). Indeed, /jlc(x) — 0 implies that x E AC for all A > 0, which by
the boundedness of C only allows for x = 0.
In (ii) we proved Hc{x) < cIMh an upper estimate follows from C C
dBx- The equivalence then implies completeness of the new norm.
2.16 Let y be a subspace of a Banach space X and || • || be an equivalent
norm on Y. Show that || • || can be extended to an equivalent norm on X.
Hint: Let B2 be the unit ball of the original norm of X. Assume without
loss of generality that the unit ball B\ of || • || on Y contains S2 fl Y. The
Minkowski functional of the set conv(jBi U B2) yields the desired norm.
2.17 Let y be a subspace of a Banach space X. Show that there exists a
one-to-one (in general, nonlinear) isometric map <p: Y* —*• X*.
Also, X* |y = y* (X* |y is the set of restrictions to Y of all / E X*).
Hint: Use the Hahn-Banach theorem to extend functionals on Y.
2.18 Show that if Y is a subspace of a Banach space X and X* is separable,
then so is Y*.
Hint: y* is isomorphic to the separable space X*/Y1.
2.19 Show that l\ is not isomorphic to a subspace of Cq.
Hint: The dual of H\ is nonseparable.
2.20 Let X be a Banach space.
(i) Show that in X* we have XL = {0} and {0}1 = X*. Show that in X
we have (X*)x = {0} and {0}_l = X.
(ii) Let A C B be subsets of X. Show that BL is a subspace of AL.
Hint: Follows from the definition.
2.21 Let X be a Banach space. Show that:
(i) span(^) = (A1)! for A C X.
(ii) span(B) C (B^I- for 5 C X*. Note that in general we cannot put
equality.
(iii) AL = ((^-L)x)"L for A C X and ?1 = {{BL)L)± for 5 C X*.
Hint: (i): Using the definition, show that A C (A1-)^. Then use that B± is
a closed subspace for any B C X*, proving that span(A) C (^"^i. Take
any x ? span(A). Since span (A) is a closed subspace, by the separation
theorem there is / E X* such that f(x) > 0 and /|___. . — 0. But then
f\ = 0, hence / E AL\ also /(a:) > 0, so x ? {A1-)^.

56 2. Hahn-Banach and Banach Open Mapping Theorems
(ii): Similar to (i).
(iii): Applying (i) to AL, we get AL C {{AL)x_) . On the other hand,
using A C (AL)± and the previous exercise, we get ((AL)±_) C AL. The
dual statement is proved in the same way.
2.22 Let X - R2 with the norm ||;r|| = (|zi|4 + |z2|4)*- Calculate directly
the dual norm on X* using the Lagrange multipliers.
Hint: The dual norm of (a, b) G X* is sup{a#i + bx2] x\ + x\ — 1}. Define
F(xi,X2,\) = olx\ + bx2 — A(x| + x\ — 1) and multiply by x\ and #2,
respectively, the equations you get from ^- ~ 0 and j~ = 0.
2.23 Let T be a set and let p G [1, oo), q G A, oo] be such that ? + ^ = 1.
Show that c0(r)* = l^T) and lp(Vy = iq(T).
Hint: See the proofs of Propositions 2.14, 2.15, and 2.16.
2.24 Show that c* is isometric to l\.
Hint: We observe that c = cq 0 span{e}, where e = A,1,...) (express
x — {?i) G c in the form x — ^e + xo with ?o — lim(?2) and xq G
i—>oo
Co). If u G c*, put vf0 = w(e) and ^ = u(e2) for z > 1. Then we have
oo
u{x) - ufac) + u(xq) = ^o^o + E u*(& ~ fo) and (vi, v2,...) G 4 as in
* = 1
oo
Proposition 2.14. Put w = (vo, wi, •..), where u0 = i>q — E ufj and write
2 = 1
OO
? = (fo,6, • • •)¦ We nave w(*0 - ^o^o + E v*& - fi(*)-
1=1
Conversely, if ii G ^i, then the above rule gives a continuous linear
functional u on c with ||w|| < ||fi[|, because \u{x)\ < (E \vi\\ sup|?;| =
\-=o ' *>o
||fi||sup |&| = ||fi||i||^||oo- The inequality ||u|| < ||u|| follows like this:
*>o
Let & be such that \v{\ = ?{Vi if V{ ^ 0 and & = 1 otherwise,
i = 0,1,.... Set xn = (?i,-.-,?n,&,&,...)¦ Then Hx"^ = 1 and
oo
|w(a;n)| = \u(xn)\ > \v0\ + ? kl - ? H- Since \u(xn)\ < ||u||, we
z = l «=n-f-l
n oo
have ||u|| > |vo| + E \vi\~ E K'l- % letting n —> oo, we get ||?i|| < \\u\\.
i=l i=n + l
2.25 Let p G A, oo) and Xn be Banach spaces for n G N. By X = (E Xn)
we denote the normed linear space of all sequences x = {^z}^ij %i ? Xj,
such that E ||^z|lx < oo, with the norm ||ar|| = (E \\x{\\px.) p.
Show that X is a Banach space and that X* is isometric to (E^-7)
(where - -f- ^ = 1) in the following sense: to / G X* we assign {/i}^ such
that /f G X* and /({ar,-},^) = ?/*(*,-).
Note: This direct sum is sometimes also denoted ® X(.

2. Hahn-Banach and Banach Open Mapping Theorems 57
Hint: Follow the proof for lp, which is the case of X{ — R.
2.26 We proved the closed graph theorem using the open mapping theorem.
Now prove the open mapping principle using the closed graph theorem.
Hint: First, prove it for one-to-one maps using the fact that {(?/, T~1(y))}
is closed. For a general case, note that the quotient map is an open map
by the definition of the quotient topology.
2.27 Let X, Y be normed spaces, T G B(X, Y). Show that f: Xj Ker(T) ->
Y defined by T(x) — T(x) is a bounded linear operator onto T(X).
2.28 (i) Prove directly that if X is a Banach space and / is a nonzero
linear functional on X, then / is an open map from X onto the scalars.
(ii) Let the operator T from c0 into cq be defined by T[(x{)) — (jX{). Is
T a bounded linear operator? Is T an open map? Does T map cq onto a
dense subset in Co?
Hint: (i): If f(x) = 6 > 0 for some xeB°, then (-<5, 6) C f(B$).
(ii): Yes. No. Yes (use finitely supported vectors).
2.29 Let T be a linear operator (not necessarily bounded) from a normed
space X into a normed space Y. Show that the following are equivalent:
(i) T is an open map.
(ii) There is 6 > 0 such that 6BY C T{BX).
(iii) There is M > 0 such that for every y G Y there is x G T~1(y) satisfying
M\x < Af||y||y.
Hint: (i) => (ii): T(B%) is open, so y0 + <$#? C T(B%) for some yo G Y,
6 > 0. T{B<x) is symmetric and convex, hence every 6 G <S?y belongs to
T(B%) since 6 = |((ar0 + 6) - (x0 - b)).
(ii) => (i): T(B^) D T(^Bx) D \BY D f 5^, so T maps neighborhoods
of 0 to neighborhoods of 0. By linearity, T is open.
2.30 Let X, Y be normed spaces, T G B(X, Y). Show that if X is complete
and T is an open map, then Y is complete.
Hint: Use (iii) in the previous exercise and Exercise 1.17.
2.31 Let X, Y be Banach spaces, T G B(X, Y). Show that if T is one-to-one
and By C T(Bx) C J3y, then T is an isometry onto Y.
Hint: Since J9y C T{Bx)) T is onto (Exercise 2.29) and hence invertible.
From T(Bx) C By we get ||T|| < 1. Assume that there is x G Sx such that
||T(x)|| < ||x||. Pick 6 > 1 such that S\\T(x)\\ < 1. Then T(Sx) E^C
T(BX)- Thus, there must be z G 5j such that T(z) = TFx), but it cannot
be 8x ? Bxi a contradiction with T being one-to-one.
2.32 Let X, Y be Banach spaces and T G B{X, Y). Show that the following
are equivalent:
(i) T(X) is closed.

58 2. Hahn-Banach and Banach Open Mapping Theorems
(ii) T is an open map when considered as a map from X onto T{X).
(iii) There is M > 0 such that for every y G T(X) there is x G T~1(y)
satisfying \\x\\x < M||y||y.
Hint: (i) => (ii): Theorem 2.24. (ii) => (iii): Exercise 2.29. (iii) => (i):
f :X/Ker(T) -> Y satisfies ||f (x)\\ > jf\\x\\] use Exercise 1.27.
2.33 Let X,Y be Banach spaces and T G B(X,Y). Show that if T maps
bounded closed sets in X onto closed sets in Y, then T(X) is closed in Y.
Hint: Assume T(xn) -> y ? T(X). Put M = Ker(T), set dn = dist(sn, M),
and find wn E M such that c?n < ||#n — wn\\ < 2dn. If {zn — wn} is
bounded, then T(xn — wn) —± y G r(X), since the closure of {xn — wn}
is mapped onto a closed set containing y. If ||arn — wn\\ —> oo, then since
T{xn - wn) -> y, we have r(,,^~%%) -+ 0. By the hypothesis, M must
contain a point w from the closure of {n^nI^n|i} since 0 lies in the closure
of the image of this sequence. Fix n so that |||j^n~^n,, — w\\ < 1/3. Then
H^n—^-|kn—Wn||w|| < |||a:n-«;n|| < B/3)dn and iun+||a:n—iun||iu G M,
a contradiction.
2.34 Let T G B(X,Y). Prove the following:
(i) KerjT) = T*(Y*)X and Ker(T*) = T(XI.
(ii) T(X) = Ker(T*)j_ and T*(Y*) C Ker(TI.
Hint: (i): Assume x G T*(Y*)j_. Then for any # G Y* we have g(T(x)) =
T*(g)(s) = 0, and hence T(z) = 0. Thus x G Ker(T).
(ii): T(X) = span(r(X)) = (T^U - Ker(T*)±.
2.35 Let X,Y be normed spaces, T G B(X,Y). Consider f(x) = T(x)
as an operator from X/Ker(T) into T(X). Then we get f*:T(X)* -±
(X/Ker(T))*. Using Proposition 2.7 and T(XI = T(X)-1 = Ker(T*), we
may assume that T* is a bounded linear operator from Y*/Ker(T*) into
Ker(TI C X*. On the other hand, for T*:Y* -+ X* we may consider
f*: Y*/Ker(T*) -+ X*. Show that f* = T*.
Hint: Take any y G Y*/Ker(T*) and x G X. Then, using the above
identifications, we obtain
f*(y*)(x) = ^ (f (*)) = y*(T(x)) = T*(y*)(x) = f=(if )(*).
2.36 Let X, Y be Banach spaces and T G #(X, Y). Show that T maps X
onto a dense set in Y if and only if T* maps Y* one-to-one into X*.
Also, if T* maps onto a dense set, then T is one-to-one.
Hint: If T(X) / Y, let / G Y* \ {0} be such that / = 0 on T(X). Then
T*(/) = 0. The other implications are straightforward.
2.37 Let X,Y be Banach spaces and T G #(X, Y). If T is one-to-one, is
T* necessarily onto?

2. Hahn-Banach and Banach Open Mapping Theorems 59
Hint: No, consider the identity map from l\ into ?2.
2.38 Let X, Y be Banach spaces and T G B(X, Y). If T is an isomorphism
into Y, is T* necessarily an isomorphism into X*?
Hint: No, embed R into R2.
2.39 Let X, Y be Banach spaces and T G B{X, Y). Show that:
(i) T* is onto if and only if T is an isomorphism into Y.
(ii) T is onto if and only if T* is an isomorphism into X*.
(iii) T(X) is closed in Y if and only if T*(Y*) is closed in X\
Hint: (i): If T* is onto, it is an open map (Theorem 2.24), and by
Exercise 2.29, there is 6 > 0 so that 8BX* C T*(BY*). Then
\\T(x)\\Y = sup y*(T(x))= sup T*(y* )(*) = sup (**(*))
y*?BY* y*EBY* x*?T*(BY*)
> sup (x*(ar)) =?||z||;r
and use Exercise 1.27.
If T is an isomorphism into, then T is a bounded linear operator from
T(X) into X. Given z* G X*, define </* on T(X) by y*(y) = ^(T^)).
Clearly, y* G T(X)*; extend it to a functional in Y*. Then T*(y*) = 3*.
(ii): If T is onto, as in (i) we find 6 > 0 such that 6By C T(BX)\ then
||T*(y*)||x* > %*||y* and use Exercise 1.27.
Assume T* is an isomorphism into. By Exercise 2.29 and Lemma 2.23,
it is enough to find 6 > 0 so that 6By C T{BX). Assume by contradiction
that no such 6 exists. Then find yn —> 0 such that yn ? T(BX). The set is
closed, so dn = dist(yn,T(Bx)) > 0.
Fix n, and set Vn — (J (y + By D^)). Then V^ is an open convex
yeT(Bx)
set and yn ^ Vni so by Corollary 2.13 there is y* G Y* such that \y*\ < 1
on Vn and y*(y„) = 1. Since T(BX) C 7n, we get
||T*(y*)|| = sup T*(if )(*) = sup y*(T(z)) - sup (</*(</)) < 1,
*G#x xeBx yeT(Bx)
so ||y*n < iicn-MIIITWH < OCT*)!!, 1 = y*(y„) < IKT*)-1!! \\vn\\.
This shows that ||yn|| > 1/||(T*) :|| for every n, contradicting yn —> 0.
(iii): If T(X) is closed, then T is an operator from Xj Ker(T) onto a
Banach space T(X), hence by (ii), T* is an isomorphism into, and in particular
f*(Y*/Ker(T*)) is closed. By Exercise 2.35, f*(Y*/Kei(T*)) = T*(Y*)
is closed.
If T*(Y*) is closed, consider f:X -> T(X). Then f*(Y*/Ker(T*)) =
T*(Y*/Ker(T*)) = T*(Y*) is closed and f* is one-to-one; hence it is an
isomorphism into. By (ii), T must be onto, that is, T(X) = T(X).
2.40 Show that there is no T G #(^2>^i) such that T is an onto map.

60 2. Hahn-Banach and Banach Open Mapping Theorems
Hint: T* would be an isomorphism of i^ into ?2, which is impossible since
^oo is nonseparable and ?2 is separable.
2.41 Let X, Y be Banach spaces, T E B(X,Y). Show that:
(i) T is an isomorphism of X onto Y if and only if T* is an isomorphism
ofY* ontoX*.
(ii) T is an isometry of X onto Y if and only if T* is an isometry of Y*
onto X*.
Hint: (i): Follows from Exercise 2.39.
(ii): If T is an isometry, then by (i), T* is an isomorphism. Also, T(Bx) —
By, so ||T*(y*)|| = sup T*(y*)(x) = \\y*\\. The other direction is similar.
xeBx
2.42 Show that cq is not isomorphic to C[0,1].
Hint: Check the separability of their duals.
2.43 We have ||T|| = ||T*|| for a bounded linear operator on a Banach
space, so if for a sequence of operators Tn we have ||Tn|| —> 0, then \\T„\\ —»
0. Find an example of a sequence of operators Tn on a Banach space X such
that ||Tn(s)|| -> 0 for every z E X but it is not true that ||T*(a:*)|| -+ 0 for
every z* E X*.
Hint: Let Tn(x) = (zn,zn+i,...) in^2- Then T*(x) = @,... ,0, xi5x2,...),
where x\ is on the n-th place.
2.44 Let X be a normed space with two norms || • ||^ and || • H2 such that
X in both of them is a complete space. Assume that || • ||x is not equivalent
to || • ||2. Let I\ be the identity map from (X, || • 1^) onto (X, || • ||2) and I2
be the identity map from (X, || • ||2) onto (X, || • Id). Show that neither I\
nor I2 are continuous.
Hint: The Banach open mapping theorem.
2.45 Let L be a closed subset of a compact space K. Show that C(L) is
isomorphic to a quotient of C(K).
Hint: Let Q: C(K) -> C{L) be defined for / E C(K) by Q(f) = f\L. Then
Q is onto by Tietze's theorem; then use Corollary 2.25.
2.46 Let X,Y be Banach spaces and T E B(X,Y). Show that if Y is
separable and T is onto Y, then there is a separable closed subspace Z of
X such that T(Z) = Y.
Hint: Let {yn} be dense in By, and take xn E X such that T(xn) — yn and
||a?n|| < K for some K > 0 (Corollary 2.25). Set Z = span{xn}; clearly,
T(Z) C Y. By the density of {yn}, B$ C T(KB%)\ hence, by Lemma 2.23,
we have B$ C T(KB°). Thus Y C T(Z).

2. Hahn-Banach and Banach Open Mapping Theorems 61
2.47 Let Y be a closed subspace of a Banach space X. Assume that X/Y
is separable. Denote by q the canonical quotient map of X onto X/Y. Show
that there is a separable closed subspace Z C X such that q(Z) =• X/Y.
Hint: Apply the previous exercise.
2.48 Let X be a Banach space, and let Y be a separable closed subspace
of X*. Then there is a separable closed subspace Z C X such that Y is
isometric to a subspace of Z*.
Hint: Let {/n} be dense in Sy* • For every n, let {#? }& C Sx be such that
fn(Xn) —>¦ 1 as fc —> oo. Put Z = span{s*; n, fe G N}.
2.49 Let X be the normed space of all real-valued functions on [0,1] with
a continuous derivative, endowed with the supremum norm. Define a linear
map T from X into C[0,1] by T(f) — /'. Show that T has a closed graph.
Prove that T is not bounded. Explain why the closed graph theorem cannot
be used here.
Hint: The graph of T is closed: let (/„,/„) -> (/, g) in X 0 C[0,1]. Then
fn —> f uniformly on [0,1]; hence, by a standard result of real analysis,
fn —> f uniformly on [0,1]. By uniqueness of the limit, /' = g.
T is not bounded: use {fn} bounded with {/^} unbounded. The space
in question is not complete.
2.50 Let X be a closed subspace of C[0,1] such that every element of X
is a continuously differentiate function on [0,1]. Show that X is finite-
dimensional.
Hint: Let T:X -+ C[0,1] be defined for / G X by T(f) = /'. Show that
the graph of T is closed: If fn —> / uniformly and fn —* g uniformly, then
ff = g. Therefore T is continuous by the closed graph theorem.
Thus for some n G N we have H/'Hoo < n whenever / G X satisfies
||/||oo < 1. Let xi — —¦ for i = 0,1,..., 4n. Define an operator S: X —»
R4n+1 by S(f) — {f{xi)}- We claim that S is one-to-one. It is enough to
show that if ||/||oo = 13 then for some f, S(f)(xi) ^ 0. Assume that this is
not true. If f(x) — 1 and x G (^, -^), then by the Lagrange mean value
theorem we have \f(x) - f(^)\ = \f(?)\\x - ^| < n • ^, a contradiction.
Therefore dim(X) < 4n+ 1.
2.51 (Grothendieck) Let X be a closed subspace of I,2[0,l] whose every
element belongs also to Loo[0,1]. Show that dim(X) < co.
Hint: The identity map from X to (?oo[0,1], || • H^) has a closed graph,
so for some a we get ||/||oo < ^H/lb for every f e X. Let {/i,...,/n}
be an orthonormal set in X. For every z = {#i,...#n} G Cn, we put
/r = J2xkfk> Then !/*(*)! < a||M|2 < a||ar||2 for almost all t G [0,1], so if
A is a countable dense set in Cn, there exists a set of measure zero N such
that |/xC0l ^ alkll2 for every ar G A and every t G [0, l]\iV. Each mapping

62 2. Hahn-Banach and Banach Open Mapping Theorems
x »-» fx(t) from Cn to C is linear and continuous, so |/rr(^)| < ^||^||2 for
all x G Cn and * G [0,1] \ TV. In particular, l/^OOl < a for z G Bc» and
t G [0,1] \ N. The choice * = (^(t),..., /n(t)) gives us ? |/fc(*)|2 < a2.
Integration then gives n = \\ J2fk\\l — Y1I \fa{t)\2 dt < ^2-
2.52 Show that the bounded linear one-to-one map <f> from Li[0,27r] into
Co defined by T(f) = f(n), where f(n) are Fourier coefficients of /, is not
onto Co.
Hint: If T were onto Co, then by the Banach open mapping theorem, T~l
would be bounded, which is not the case, as the sequence X{i,...,n} shows
(note that we have ||Ai||i —* oo, where Dn is the Dirichlet kernel).
2.53 Show that there is a linear functional L on i^ with the following
properties:
AI1^1 = 1;
B) if x — (xi) G c, then L(x) — lim (ar8-);
i—+-oo
C) if x — (x{) G ioo and X{ > 0 for all i, then L(x) > 0;
D) if x — (x^ G ^oo and x' — (x2j x%,...), then L(x) = L(x').
Hint: For simplicity, we consider only the real scalars setting. Let M be
the subspace of l^ formed by elements x — xf for x G ^oo and x' as above.
Let 1 denote the vector A,1,...). We claim that dist(l, M) — 1. Note that
0 G M and thus dist(l,M) < 1. Let x G 4o- If (x - x')i < 0 for any of
z, then || 1 — (x — a^Hoo > 1. If (x — x')i > 0 for all z, then Xi > #2+i for
all i, meaning that lim(xz) exists. Therefore lim(^z- — x^) — 0, and thus
iii-(*-*')h>i.
By the Hahn-Banach theorem, there is L G t^ with ||L|| = 1, L(l) = 1,
and L(m) — 0 for all m G M. This functional satisfies A) and D). To
prove B), it is enough to show that Co C L-1@). To see this, for x G -^oo we
inductively define x^ — x' and #(n+1) = (#(n))' and note that by telescopic
argument we have a?(") -xeM. Hence, L(x) = L(x^) for every x G -too
and every n. If x G Co, then ||#(n)|| —> 0 and thus L(x) = 0. To show C),
assume that for some x = (xn) we have x\ > 0 for all i and L(x) < 0. By
scaling, we may assume that 1 > X{ > 0 for all i. Then J|l — a?||oo < 1 and
L{\ — x) — 1 — L(x) > 1, a contradiction with ||L|| = 1.

3
Weak Topologies
Given a normed space (X, || • ||), by X** we denote the space (X*)* with
the norm ||.F|| = sup \F(f)\. We define higher duals by induction as
feBx.
X*** = (X**)*,etc.
Definition 3.1
Let X be a normed space. The canonical embedding it of X into X** is
defined for x G X by
*(x)'f*->f(x).
Note that w is a linear isometry. Indeed,
*(<xx + Py)(f) = f(ax + Py) = af(x) + /3f(y)
= Mx) + ^y)](/).
Also, for x G X we have ||7r(x)|| = sup |/(x)| < sup ||/|| • ||ar|| < ||a:||.
f?Bx* feBx*
Considering /o G Sx* such that fo(x) — ||#||, we have ||7r(^)|| > |/o(^)| =
|\x11. Therefore ||fl-(a;)|| = ||ar||.
Using this natural identification, for x G X we often write x G X**
instead of tt(x) G X**: and we identify X with ir(X) C X**. In particular,
x(f) = *(/) for x G X and / G X*.
Fact 3.2
For every normed space X, there exists its completion X, that is, a Banach
space X such that X is a dense subset of X.

64 3. Weak Topologies
Proof: We can use X = 7r(X)
D
Given normed spaces X, Y and T E B(X,Y), we similarly define T** =
(T*)*. Note that for z E X we have T**(v(x)) = 7r(T(x)); in particular,
T**(tt(X)) C 7r(y). We have T**(X) C y and T**|x = T.
Definition 3.3
Let X be a normed space.
The weak (w-) topology on X is the topology generated by a basis consisting
of the sets
O = {x E X; \fi(x - x0)\ < e for i = 1,..., n}
/or all choices of xo ? X, /i,...,/nEl* and ? > 0.
Similarly, the weak star (w*-) topology on the dual X* of X is generated
by a basis consisting of the sets
0* = {feX*; \(f-fo){xt)\<efori=l,...,n}
for all choices of fo E X*, xi,...,xnEl and e > 0.
In the complex case, we may equivalently define the tu-topology on X
using only functionals from Xr,. The iu*-topology on X* may also be
equivalently defined using | Re(/ - fo)(x{)\ in the description of its basis.
Let X be a real Banach space. By a half space of X we mean a weakly
open set of the form {x E X] f(x) < a} for some / E X* \ {0} and a E R.
Finite intersections of half spaces form a basis for the weak topology.
Note that the w*- and w-topologies are Hausdorff. Indeed, given / ^ g in
X*, there is x E X and a scalar a such that f(x) > a > g(x). The tu*-open
sets {h E X*; /i(z) > a] and {ft E X*; ft(x) < a) separate / from g. The
proof for the tu-topology is similar using the Hahn-Banach theorem. Note
also that if a set A C X is tu-open (resp. A C X* is w*-open), then the set
x + A is w-open (resp. w*-open) for any x E X (resp. ? E X*).
It is immediate that every w-open set is also norm open, so the norm
topology is stronger than the weak topology. Since the sets defining the
u?*-topology on X* are among the sets defining the w-topology on X*, the
weak topology is stronger than the weak star topology on X*. Also, A C X
is w-compact if and only if tt(A) is w*-compact in X**.
We will use M and M for the closure in the respective weak
topologies. We will also use conva(M) for the <r-closed convex hull of M.
Properties of a general topological space cannot always be described
using sequences. A more general concept, nets, is needed. We will describe
here the required definitions and some simple facts about nets.
Definition 3.4
By an index set we mean any partially ordered set I; that is, a set I with
a binary relation < on I satisfying for all a,/?,7 E I:

3. Weak Topologies 65
(l)a<a;
(#) if ol < fl and ft < 7, then a < 7;
C) if a < ft and C < a, then a — /?.
Note that we do not assume that two arbitrary members of / are always
related.
Definition 3.5
A net 27i a nonempty set X is a mapping N from an index set I into X.
Instead of N(a), we will often write xa and denote the net as {xa}a^j.
Let {xa}a?i be a net. Let J be an index set and S: J —> I a mapping
with the following property: given ao G /, there exists /?o G J such that
ao < S(/3) whenever /3q < C. Then the net {xs(p)}p?j is called a subnet
of the net {xa}a?i.
Assume now that X is a topological space. We will describe convergence
of nets defined in X.
Definition 3.6
We say that a net {xa}a?i in a topological space X converges to some point
x G X if for every neighborhood U(x) of x there exists ao G I such that
xa G U(x) whenever ao < a. We then say that x is the limit of {xa}aej
and write X{ —» x.
We say that x G X is a cluster point of a net {xa}aej if for every
neighborhood U(x) of x and ao G I there exists a ? I such that ao < a and
xa G U(x).
Note that if xa —» x, then every subnet of {xa} also converges to x. A
net converges to x if and only if every subnet has a? as a cluster point.
The topology of a topological space X can be specified by describing the
convergence of nets. Hence, all topological concepts can be defined in terms
of nets. For example, given a subset A of a topological space X, its closure
A is equal to the set of all limits of nets in A. Thus, a set A is closed if and
only if it contains the limits of all convergent nets with elements in A.
Note that in a metrizable topological space X it is enough to consider
sequences. For more information on nets, we refer to [Kel].
Let us introduce some convenient notation for convergence. xn —> x (or
lim(arn) = x) will mean the convergence in norm unless stated otherwise.
We also write (sometimes w-lim(?n) = x) to say that {xn} con-
w*
verges to x in the weak topology of X\ for functionals, we use fn —» /
(sometimes u>*-lim(/n) = /) to say that fn converges to / in the weak star
topology of X*. Similar convention applies to convergence of nets.
Note that xn —> x implies xn —-> x. Similarly, in the dual space X* we
have that /«—>•/ implies /n —>• / implies fn —> /. The weak convergence
of sequences is easily characterized as follows:

66 3. Weak Topologies
Proposition 3.7
Let X be a normed space.
(i) Let f,f1,f2,...eX*. Then fn^U f if and only if lim (fn(x)) = f(x)
n—>-oo
for every x ? X.
(ii) Let x, x\, #2,. • • E X. Then xn —* x if and only if lim (f(xn)) = f(x)
n—+oo
for every f ? X*.
In the situation described in (i), we say that fn converges to f pointwise
on X. More generally, for maps F, Fn: X —> Y and a subset G C X, we say
that Fn —> F pointwise on G if Fn(x) —» F(x) (convergence in Y) for every
xeG.
Proof: (i): Assume that fn —» / pointwise on X. If O is an open set in
the w*-topology containing /, there is e > 0 and #i,..., xm EX so that
{g e X*; \(g - f)(Xi)\ < e, i = 1,..., m} c O.
Since fn(xi) —> f(xi) for all i = 1,..., m, there is no so that |(/n~/)(^)| <
e for n > no. Therefore, /n is eventually in every open set containing /.
On the other hand, let lim(/n) = / in the w*-topology and consider
n
x ? X. Given e > 0, the set A = {g ? X*; \(g — f)(x)\ < e} is ^*-open,
and thus for n large enough we have fn ? A, meaning \fn(x) — f(x)\ < ?-
(ii): The proof is similar.
D
In the exercises, we will show that for fn ^-> /, where {/n} is a bounded
sequence in X*, it is enough to check fn(x) —> f(x) for x belonging to some
set M such that span(M) = X.
We will also show there that if xn —> x) then {xn} is bounded and
||#|| < liminf \\xn\\, and analogously for the tu*-convergence.
Proposition 3.8
// a normed space X is finite-dimensional, then the weak topology of X
coincides with the norm topology of X, and the weak star topology of X*
coincides with the norm topology of X*.
Proof: We prove the latter fact by showing that every open set in X*
is tu*-open. To see this, let O C X* be open. Let fo € O and e > 0 be
such that /o + eBx* C O. Let {ei,...., en} be a basis of X, and we define
l/l = max |/(e,-)|. Since all norms on X* are equivalent (X* is finite-
l<z'<n
dimensional), we get that there is 8 > 0 such that for every / ? X*, ||/| <
6, we have ||/|| < e. Then the w*-open set {/; max|/(e2) — /o(e2)| < 6} is
contained in {/; ||/ — /o|| < ?}• Thus, O is w*-open.
?

3. Weak Topologies 67
For infinite-dimensional spaces X, the iu*-topology never coincides with
the norm topology of X*. To prove this latter fact, we first need the
following lemma.
Lemma 3.9
Let X be a vector space, and let /, /i, fi,. •., fn be linear functional on X.
n
V n /z~1(^) C /_1@); 'then f is a linear combination of fi,..., fn.
We use the following fact from linear algebra.
Claim
Let Ei,E2,Es be vector spaces, and let f'.E\ —» E3 and g\E\ —> E2 be
linear mappings. There is a linear map h: E2 —» #3 sucA tfAfltf f — h o g if
and only ifg'1^) C /_1@)'
Proof: Suppose that g^) C /_1@)- Define ^:g[Ei] -+ E3 by h(g(x)) =
/(x) for x G ?". To check the consistency, assume g(x\) = g(x2). Then
(#i — #2) € 5r_1@) C /-1@), so /(ari) = f(x2). Extend h to a linear map
on ?2, clearly h(g) = /. The other implication is clear.
?
Proof of Lemma 3.9: Using the Claim with Ei = X, E2 - Rn, ?3 = R,
/ = /, and </(#) = (/i(a?),. • • ,/n(#)), we get that there is a linear map
/i:Rn —> R such that f(x) = h(g(x)). The mapping h can be written as
h{y) — Y2aiVi f°r some c*i,... ,an and every y = (yi) G Rn. Therefore,
f(x) - 12aifi(x) for every x ? x.
?
Proposition 3.10
Zetf X fee an infinite-dimensional normed space and O C X. If O ^ $ is
w-open, then O is not bounded.
In particular, the w- and norm topologies do not coincide.
Proof: We can assume that 0 G O. Then there is e > 0, /1,..., /„ G X*
such that {x] \fi(x)\ < e} C O. Clearly, the set
N = {x G X; fi(x) = 0 for i = 1,..., n} = p| fr\0)
is contained in O. We claim that N ^ {0}. Indeed, suppose that TV = {0}.
Then for every / G X* we have N C /-1@), so by Lemma 3.9, / is a linear
combination of /1,..., fn. Thus X* = span{/2}, a contradiction.
Therefore we can find 0 ^ x G N. Then for every scalar A we have
Xx G N, so O contains a line through x. Hence O cannot be bounded in
X.
D

68 3. Weak Topologies
Definition 3.11
Let X,Y be normed spaces and T C B{X)Y). We say that T is pointwise
bounded z/sup{||T(z)||y; T E F) < oo for all x eX.
If T is bounded in B(X,Y)—that is, there is C > 0 so that ||T|| < C
for all T E T—then T is pointwise bounded. Indeed, for x E X we have
\\T(x)\\y < \\T\\\\x\\x < C\\x\\x, so sup{||T(x)||y; T € F} < C|M|jc
The opposite direction is often called the (Banach-Steinhaus) Uniform
boundedness principle.
Theorem 3.12 (Banach, Steinhaus)
Let X, Y be Banach spaces and T C B(X, Y). If T is pointwise bounded,
then T is bounded in B(X,Y).
Proof: For n E N, set Nn = {x E X\ sup ||T(x)||y < n}.
We claim that Nn is closed, convex, and symmetric in X. The symmetry
of Nn is obvious. To check closedness, let Xk E Nn and x^ —> x E X. Given
T E T, we have ||r(xjb)||y < n, so ||T(z)||y < n by continuity. To see that
Nn is convex, let Xi,x2 6 Nn and A E [0,1]. Then, for every T G f, we
have
\\T(Xx1^(l-X)x2)\\Y < X\\T(x1)\\y + A-X)\\T(x2)\\y < An+A-A)n = n.
Since for every x ? X we have sup ||T(a?)||y < oo, there is some n E N
OO
greater than the supremum. Hence x E Nn, so (J 7Vn = X. By the Baire
n = l
category theorem, there is no such that the set Nno contains an interior
point xq. Thus, there is S > 0 such that xq + 6Bx C iVno. Because of the
symmetry of JVno, we have — x0 + 6Bx C Nno. If 6 E Bx, then by the
convexity of Nno we have b = |(a?o + b) + |(—a?o + 6) E iVno. Hence SBx C
ATno. Consequently, given T ? J7, for every a? E Bx we have ||TF^)||y < no;
that is, ||T|| < *f. This means that sup ||T|| < ?f.
D
Corollary 3.13
Let XyY be Banach spaces and Tn E B(X,Y) for n E N. Assume tta^ /or
even/ a? E X tfiere exists the limit T(x) = lim (Tn(x)). T/ien T E B(X, Y)
and ||r||<liminf||T„||.
n—»-oo
Proof: The linearity of T is easy:
T(x + y) = lim(r„(z + 2/)) = lim(Tn(x) + Tn(y))
= lim(Tn (a:)) + lim(Tn (y)) = T(x) + T{y).
From Tn(z) —> T(z) we have ||Tn(z)|| —> ||T(x)||; in particular, we get that
{||Tn(ar)||; n E N} is bounded for all x E X. By the Banach-Steinhaus

3. Weak Topologies 69
theorem, {Tn} is bounded in B(X,Y). Denote C = liminf ||Tn||. Then
||T(x)|| = lim||Tn(z)|| = liminf ||Tn(s)|| < liminf(||x|| ||Tn||).
This shows that \\T{x)\\ < C\\x\\\ that is, T is bounded and ||T|| < C.
D
Definition 3.14
Let X be a Banach space. We say that a set M C X is weakly (w-) bounded
if sup{|/(a:)|; x G M} < oo for every f G X*.
We say that a set M C X* is w*-bounded ifsup{\f(x)\) f G M} < oo for
every x G X.
Theorem 3.15 (Banach, Steinhaus)
Let X be a Banach space. If M C X* is w*-bounded then M is bounded.
IfMcX is w-bounded then M is bounded.
Proof: If M C X* is w*-bounded, then M is pointwise bounded in
B(X, K); hence, by Theorem 3.12, it is bounded in B(X, K) = X*.
If M C X is tu-bounded, then 7r(M) is pointwise bounded in B(X*, K);
hence, by Theorem 3.12, ir(M) is bounded in B(X*,K) = X**. Since it is
an isometry, M is bounded in X.
D
Theorem 3.16
Let X be a Banach space and F G X**. If F is continuous in the weak star
topology, then there is x G X such that F — tt{x) {or, briefly, F G X).
Note that by definition of the w*-topology, every tt(x) G X** is w*-
continuous.
Proof: If F is ^-continuous on X*, then it is bounded by one on some w*-
neighborhood U of 0 in X* of the form U — {f E X* ] max |/(zt-)| < e},
l<i<n
where #i,..., xn EX and e > 0. Let / G X* be such that f(xi) = 0 for
i = 1,..., n (Exercise 2.11). Then kf G U, so |F(Jb/)| < 1 for every jfc. Thus
F(f) - 0. Therefore, f) ^(a;,-)^) C F-^Q); hence, by Lemma 3.9, F is
2 = 1
a linear combination of tt(^i), ..., Tr(xn). Thus F € X.
D
We will now prove more separation theorems. For simplicity, we state
only real versions.
Theorem 3.17
Let A,B be disjoint closed convex subsets of a real Banach space X. If A
is weakly compact, then there is f G X* such that supA(/) < infs(/).
Proof: There is a weak neighborhood U of the origin in X such that
(U + A) fl B — 0. Indeed, every point x G A can be separated from B by

70 3. Weak Topologies
some / due to Theorem 2.12, yielding a w-open neighborhood Ux of 0 such
that (x-\-Ux)C\B = 0. By weak compactness, we cover A with finitely many
such open sets, thus obtaining U.
The set (A + U) is open as a union of open sets; hence, (^4 + U) — B is
an open set not containing the origin. Using Corollary 2.13 for this set and
the origin, we find / ? X* such that /(a + u — b) > 0 for all a ? A, b ? B,
and u eU. Thus /(a) > f(b) + supt/(/) for all a ? A, 6 ? B. Since i7 is a
neighborhood of 0, we have supc/(/) > 0 and the statement follows.
?
Theorem 3.18
Let X be a real Banach space. If A is a w*-closed convex set in X* and
f ? X* \ A, then there is x ? X such that f{x) > sup{#(#); g ? A}.
Proof: Since A is tt>*-closed, there is a ^-neighborhood U of zero such
that (/ + U) fl A = 0. We may assume that U is a convex neighborhood
of zero of the form U = {y* ? X*] \y*(xi)\ < e for i = 1,..., n} for some
xi,... ,xn ? X and e > 0.
By the symmetry of G, we get / ^ (^4 +17). Since A + U is w*-open, it is
open, also convex, and thus by Corollary 2.13 there is F ? X** such that
F(/)> sup (F(y))>sup(F(y)).
geA+U geA
We claim that F — ir(x) for some x ? X. Fix some ^oEi and note
that C = s\ipu(F) < F(f) — F(go) is finite. Consider the points X{ as
linear functional on X*. Let y* ? flar,^). Then ty* ? {/ for all * > 0,
and hence also F(ty*) < C; that is, F(y) < f. Similarly, F(-y) < f.
Consequently, F(y) = 0, and by Lemma 3.9, F is a linear combination of
zt-; that is, F ? X.
?
Theorem 3.19 (Mazur)
Every closed convex set in a Banach space X is weakly closed.
Therefore, for convex sets, the norm and the weak closures coincide.
Proof: Assume without loss of generality that X is a real Banach space,
and C is a closed convex subset of X. If xo ? C, choose / ? X* such
that f(xo) > sup{/(#); x ? C} by Theorem 2.12. Let a ? R be such that
sup{/(#); x ? C) < a < f(x0). Then O - {x ? X; /(z) > a} is weakly
open in X. Clearly, x0 ? (9 and 0flC — 0. This shows that X\C is w-open
and therefore C is unclosed.
Let C be a convex set. Since the norm topology is stronger, we obtain
CCC. Since C as a closed convex set is also w-closed, we have C = C .
?
For non-convex sets, the norm and weak closure may differ, see
Exercise 3.8.

3. Weak Topologies 71
In infinite-dimensional spaces, xn —* x does not in general imply xn —> x
in norm (for instance, e2- ^ 0 in c0 or ?p, p > 1). However, we have the
following corollary.
Corollary 3.20
Let X be a Banach space and x} x\,X2, ...GX. If x<n —> z? 2Aen there exist
convex combinations yk of {xn} such that yk —» a:.
In other words, if xn —> a:, then for every ? > 0 there are Ai ... Am > 0
such that Y^ An — 1 and \\x — ^Anzn|| < e.
Proof: Define C = conv{zn}. Since #n -^> x, we have x ? C — C,
because C is convex and closed. From ? ? conv{#n}, the statement follows.
D
If C is a subset of a Banach space X, by (C, u>) we denote the
topological space C endowed with the restriction of the ^-topology to C. Similar
notation applies to (C, w*) in a dual space.
Theorem 3.21 (Alaoglu)
Let X be a Banach space. Then Bx* is compact in the w*-topology.
Proof: By Tychonoff's theorem, the space [—1, l]Bx of all real functions
on Bx with values in [—1,1] is a compact topological space in the point-
wise topology. Consider the set Bx* \bx °f aU restrictions of / ? Bx* onto
Bx, which is clearly a subset of [—1, l]Bx. It is easy to see that the
restriction map is a homeomorphism of (Bx*, w*) onto Bx* \bx m the pointwise
topology, so for w*-compactness of Bx*, it is enough to show that Bx* \bx
is a pointwise closed subset.
Let {fa}aeA be a net in Bx* \B which converges pointwise to some
F ? [-1,1]B*. Define /@) - 0 and f(x) = |M|F(fl%) for x ? X \ {0}.
Since F is homogeneous on Bx, we also have /(z) — ^(f) for any 8 > \\x\\.
Using this and the linearity of F on Bx, we easily show that / is linear on
X. Since fa—*F pointwise and ||/a|| < 1, we have ||/|| = sup |F(a?)| < 1.
xeBx
Thus, / ? Bx* and F = f\ ? Bx* \B , as needed.
?
Proposition 3.22
Let X be a separable Banach space and let {xn}^L1 C Sx be dense in Sx•
Define a metric on Bx* by
oo
p(f,g) = 'El2-i\(f-g)(xi)\.
« = 1
TAerc 2/ie formal identity map I:(Bx* ,w*) —> (Bx*, p) is a
homeomorphism. In particular, (Bx*, p) w a compact metric space.

72 3. Weak Topologies
Proof: Clearly, p is a metric. By Alaoglu's theorem, (Bx*: w*) is compact.
We need only show that I:(Bx*,w*) —> (Bx*,p) is continuous because
a continuous one-to-one map from a compact topological space onto a
Hausdorff topological space is a homeomorphism.
Consider an open neighborhood O — {g G Bx*; p(/, g) < s} of some
/ G Bx* • There exists n0 G N such that 2"n° = ? 2"no-' < f. For every
»=i
g G -Bx* satisfying |(<? — /)(#;)| < ^~ f°r every i = 1,..., no, we then have
oo
p(ff)/) = ?2-"|(/-jO(*,-)I
2 = 1
no oo
= XV*'l(/-ff)(*.-)l + ? a-'KZ-ffKxOI
i=l 2=n0+l
oo
2 f-f
2 = 1
Thus, we have verified that O = I" ((9) contains some w*-open
neighborhood of /. Consequently, the mapping / is continuous.
?
Thus if X is separable, then the topological space (Bx*,w*) is
metrizable. In fact, the converse is also true.
Lemma 3.23
Let K be a compact metric space. The Banach space C(K) is separable if
and only if K is metrizable.
Proof: IfC(K) is separable, then Bc(k)* m the w*-topology is metrizable
by Proposition 3.22. Consider the map 8:k i—> 8k, where 6k is the Dirac
functional corresponding to k. Note that 8 is one-to-one, since given k\ ^
k2 G K, the function / = dist(fc,&i) G C(K) separates 6k1 from <5&2. It is
easy to see that 6 is a homeomorphism of K onto 6(K), considered with
the w*-topology of Bc(k)*- Since <p(K) C (Bc(k)*>w*) ls metrizable, so is
K.
Assume now that K is a compact metric space. Let {xn}(^L1 be a dense
sequence in K. Let /n,m(^) = ^ — dist(a?,;rn) if dist(#,#n) < ~ and
fnm = 0 if dist(z,?n) > ^. Then the family {fn,m} together with a
constant function generates a separable algebra that separates the points of
Kj and thus its closure is C(K) by the Stone-Weierstrass theorem ([Roy]).
?
Proposition 3.24
Let X be a Banach space, (Bx*,w*) is metrizable if and only if X is
separable.

3. Weak Topologies 73
Proof: Assume that (Bx*,w*) is metrizable. Lemma 3.23 implies that
C{Bx* ,w*) is separable in its supremum norm. The mapping I:X —>
C(Bx*,w*), I(x):f h-» /(#), 1S an isometry, and hence X is separable
as a subspace of a separable metric space.
The other implication is contained in Proposition 3.22.
?
Since X* — [JnBx*, from Proposition 3.22 we also have the following
corollary.
Corollary 3.25
Let X be a Banach space. If X is separable, then X* is w*-separable.
Proposition 3.26
Let X be a Banach space. If X is w-separable, then X is separable.
Proof: Let 5 be a countable set that is weakly dense in X\ denote D —
spanE). The countable set C of rational linear combinations of S is dense
in D, and hence D is norm separable. Since D is convex and dense in X,
— —w
by Theorem 3.19, D-D = X, so C is dense in X.
?
We will now investigate (Bx, w).
Theorem 3.27 (Goldstine)
Let X be a Banach space. The w*-closure of Bx in X** is Bx**.
Proof: Since Bx C Bx** and Bx** is tiAcompact by Alaoglu's theorem,
the w*-closure of Bx in X** is contained in Bx** •
*
Denote W — Bx and assume that W ^ Bx**. Then there is x%* ?
Bx** such that x%* ? W. By Theorem 3.18 applied to X*, there is / ?
X* such that x$*(f) > sup (y**(/)). Since the supremum is positive,
y**ew
by scaling we may assume that sup (y**(f)) = 1. Since Bx C W} in
y**?W
particular ||/|| < 1. But also ||xo*ll ^ 1 and yet x^*(f) > 1, a contradiction.
?
Proposition 3.28
Let X be a Banach space. (Bx,w) is metrizable if and only if X* is
separable.
Proof: If X* is separable, we take a set {/t-} dense in Sx* • As in Propo-
oo
sition 3.22, we obtain that the metric p(x, y) = Yl ^~%\fi{x ~ 2/)I gives the
2 = 1
topology of (Bx,w).
For the other direction, let Un be a countable basis of neighborhoods of
0 in Bx in the weak topology and assume that for every n there is en > 0
and a finite set A*n in X* such that Un = {x ? Bx; \x*(x)\ < en, x* ? A* }.

74 3. Weak Topologies
Put A* = U^n- We will show that for X{ = span(A*) we have X{ = X*.
If not, choose y* G X* with cf = dist(?/*, Xf) > 0. By the Hahn-Banach
theorem, there is x** G X** with \\x**\\ = \jd such that x**(X%) = 0 and
x**(y*) = 1. The set V - {x e BX] \y (*)\ < d/2} is a weak neighborhood
of 0 in Bx and thus it must contain some Un. We will now show that this
is not the case.
Given n, by Goldstine's theorem there is x\ G Bx such that \d—y*(x\)\ —
\dx**(y*)-y*{x{)\ < d/2 and |x*(a?i)| = \dx**(x*)-x*(xi)\ < en for every
x* G A*n. Thus, |y*(xi)| > d/2 and |x*(a?i)| < en for every x* G A*. But
this means that a?i G Un and a?i ^ V", so Un (jL V.
D
Note that the topology given by p is in fact the topology of pointwise
convergence on {/,-}; that is, xn >• x if and only if lim(/a-(a:n)) = fi(x) for
all i.
Proposition 3.29
Let C be a weakly compact set in a Banach space X. If X* is w*-separable,
then (C, w) is metrizable.
In particular, if X is separable, then (C,w) is metrizable.
PROOF: If T is a countable separating set in X*, then the topology of
pointwise convergence on T is metrizable and coincides with the weak
topology on C since C is weakly compact. Such a separating set exists
if X* is w*-separable (take any w*-dense countable set).
?
Definition 3.30
A Banach space X is said to be reflexive if the mapping it in Definition 3.1
maps X onto X**.
In particular, a reflexive space X is isometric to X**. On the other hand,
there are examples of nonreflexive spaces X such that X is linearly isometric
to X**—of course, not by means of 7r (James's space J; see Chapter 6).
We will see that X is reflexive if and only if the ^-topology and the
w*-topology agree on X*.
Examples
(i) lp and Lp spaces for 1 < p < oo are reflexive. Indeed, if — + — = 1, then
?** — tq — lp in the sense that for every / G tq there is g G ip such that
f(x) = Y^XiQi f°.r every x G ?q. The action of / on x is the same as the
action of g on x and it is thus a map from ?p onto ^**. For Lp it is shown
similarly.
(ii) Co is not reflexive since Cg* = ^oo> Co is separable, and l^ is not.
(iii) C[0,1] is not reflexive. Indeed, C[0,1]** would then be separable and
thus C[0,1]* would be separable by Proposition 2.9, which is not true
(Proposition 2.20).

3. Weak Topologies 75
Theorem 3.31
A Banach space X is reflexive if and only if Bx is weakly compact.
Proof: Let X be reflexive. Then X = X** and thus Bx = Bx** is
iu*-compact in X** by Alaoglu's theorem, so it is weakly compact in X.
If Bx is weakly compact, then it is iu*-closed in X**. Since the u>*-closure
of Bx in X** is Bx** by Theorem 3.27, we get that Bx = Bx** and the
space X is reflexive.
?
Proposition 3.32
A Banach space X is reflexive if and only if X* is reflexive.
Proof: Let X be reflexive. Then X = X** and thus the tt;*-topology and
w-topology on X* coincide. Therefore, by the Alaoglu theorem, Bx* is
weakly compact and X* is reflexive by Theorem 3.31.
Let X* be reflexive. By the first part of this proof, we have that X** is
reflexive. Therefore, Bx** is w-compact. Bx is a closed and therefore
inclosed (Theorem 3.19) subset of Bx** and thus w-compact in X**. Since
the w*-topology is weaker, Bx is w*-compact in X**, so Bx is ^-compact
in X and we are done by Theorem 3.31.
?
Proposition 3.33
Let X be a reflexive Banach space. IfY is a closed subspace of X, then Y
is a reflexive Banach space.
Proof: We first observe that the weak topology of Y coincides with the
restriction of the weak topology of X to Y. Indeed, if for some net {ya} C Y
and y E Y we have that yn —> y in the weak topology of X, then by the
Hahn-Banach theorem, every / ? Y* can be extended to / G X* and
from ya -^> y in X we have f(ya) -+ f(y). Thus ya —> y weakly in Y.
The other part of the proof of the coincidence of these two topologies is
straightforward.
To complete the proof of Proposition 3.33, it suffices to observe that By
is a weakly closed subset of the weakly compact set Bx. Therefore, By is
weakly compact in Y, and thus Y is reflexive by Theorem 3.31.
?
Theorem 3.34
Let X be a Banach space. If X is separable and reflexive, then (Bx,w) is
a compact metrizable space.
Proof: Because X** — X is separable, X* is separable by Proposition 2.9.
Then Bx in the ^-topology (as Bx** in the u>*-topology) is a compact
metrizable space by Proposition 3.22.
?

76 3. Weak Topologies
Extremal Structure
Definition 3.35
Let C be a subset of a Banach space X. A point x G C is called an extreme
point of C if x is not the center of any non-degenerate line segment in C.
The set of all extreme points of C is denoted by Ext(C).
Recall that H C X is called an affine subspace of X if there is y G X and
a linear subspace Y of X such that H = y + Y. It is easy to observe that
closed affine hyperplanes (affine subspaces given by closed hyperplanes) are
sets that can be obtained as f~l(a) for / G X* and a G K.
Consider a convex set K C X. An affine subspace H of X is called a
supporting manifold for K in X if H H K ^ 0, and whenever a line segment
[x, y] C K has an interior point in i7, then [x, y] C #.
Note that if f ? X* and if there is k G if such that swpK(f) — f(k) = a,
then iJ = /_1(^) is a closed supporting manifold for if. Indeed, suppose
that [#, y] C if and for some A G @,1) we have f{Xx + A — X)y) = a. If
/(a?) < a, then f(Xx+(l-X)y) = A/(x) + (l-A)/(y) < Aa+A-A)a = a,a
contradiction. Therefore /(z) — f(y) = a, showing that for every A G [0,1]
we have /(A + A — A)y) = a; that is, [#, y] C ii.
If if is if-compact, then for a given / G X* we can always find fc as
above, that is, there are closed supporting manifolds for K in X. Recall
that, by the Hahn-Banach theorem, for every x G Bx there is a supporting
functional / of Bx at x, and hence Z A) is a supporting manifold for Bx •
Fact 3.36
Let K be a convex subset of a Banach space X. If H is a supporting
manifold for K in X such that H C\ K — {x} for some x G X, then
x G Ext (if).
Proof: Clearly, x G if. If x = \{x\ + x2) for x\ ^ x2 G if, then the
interior of the segment [?1,2:2] C if has a point in H (namely x), and
hence [x\, x2] C H C\ K, a contradiction.
D
Theorem 3.37 (Krein, Milman)
Let X be a Banach space.
(i) If K C X is w-compact and convex, then K — conv(Ext(if)).
(ii) If K C X* is w*-compact and convex, then K = conv™ (Ext(if)).
Note that we have conv(Ext(if)) = conv(Ext(if)) by the convexity of
conv(Ext(if)). We prove only (i), for the proof of (ii) is similar. In the
proof, we may assume without loss of generality that X is a real space. We
will use the following lemma.

3. Weak Topologies 77
Lemma 3.38
Let K be a w-compact convex subset of a real Banach space X. If H is a
closed supporting manifold for K, then H contains an extreme point of K.
Proof: Let M be the family of all closed supporting manifolds for K
contained in H and partially ordered by inclusion. Note that M =? 0 since
HeM.
If {Ma} is a chain in M, then f] Ma is a closed afflne subspace of X and
(f)Ma)ni< = f)(ManK)^<b,
because Ma H K is weakly compact for every a. If a segment [x, y] C K
has an interior point in f)Ma, then [xyy] C Ma for every a since Ma is
a supporting manifold for K. Therefore [x,y] C f]Ma. This means that
f]Ma is a closed supporting manifold for K.
By Zorn's lemma, there is a minimal element Mo in M. We will show
that Mo fl K is a singleton, which must be an extreme point. Assume that
there are x ^ y E M0 fl K. Then there is / G X* such that f(x) ^ /(y).
Since Mo fl K is «;-compact, by the preceding remark there is k G K fl Mo
such that /(&) = a — sup (/). Then f~l(a) is a supporting manifold for
MqCiK
MqDK. M1 — Mo n/-1(a) is an afhne subspace (nonempty since k G M'),
and we observe that it is a supporting manifold for K. Since f(x) ^ /(y),
at least one of them is not in M;; that is, M' is a strict subset of Mo, which
contradicts the minimality of Mo.
?
Proof of Theorem 3.37: Let B = caov™ (Ext(i\)). If B ± k, choose c G
K\B and find / G X* such that /(c) > supB(/) using Theorem 2.12. For
a = sup^(/), consider H — /-1(a). This is a closed supporting hyperplane
for K, so by Lemma 3.38, H contains an extreme point x of K. But x (fc B
because f(x) = a > sup5(/), a contradiction.
?
Definition 3.39
Ze2 C be a set in a Banach space X. A slice ofC is a nonempty intersection
of C with an open half space of X.
We observed that finite intersections of slices form a basis of the weak
topology.
Lemma 3.40 (Choquet)
Let C be a w-compact convex set in a Banach space X. For every x G
Ext(C), the slices of C containing x form a neighborhood base of x in the
relative weak topology of C.
Note that a similar result holds for the weak star topology of X*.

78 3. Weak Topologies
Proof: Let V be a neighborhood of x in the relative weak topology of C of
the form V — V\ fl- • -DVk, where V{ are slices of C (i.e., V{ = VjnC and V*
A;
are open half spaces in X). Then a: ? [J ((X \ K) H C). Consequently, a? ?
conv (J ((X \ K") H C) because x is an extreme point of C (Exercise 3.71).
Since the convex hull of a finite number of weakly compact convex sets is
k
weakly compact (Exercise 1.55), we have that x ? conv |J ((X \ T4) fl C).
i=l
By Theorem 2.12, there is / G I* and a E R such that f(x) > a >
sup{/(x); x E (J (X \ V5) H c\. Then the slice C 0 {x EX; /(x) > a}
2 = 1 ^ J
contains x and is contained in V.
D
Theorem 3.41 (Milman)
Let C be a weakly compact convex set in a Banach space X. If B C C is
such that conv(jB) = C, then Ext(C) C B .
Similarly, if C is a u>*-compact convex subset of X* and conv™ (B) — C,
then Ext(C) C IT .
Proof: Assume that for x E Ext(C) we have x ^ B . Then there is a
slice 5 of C that contains x ? C and such that S C\ B = 0. Thus, there
are / E X* and a E R such that /(x) > a > sup^/). By the linearity
of /, sup(/) = sup (/), so /(x) > a > sup (/). This contradicts
B conv(B) conv(B)
C = conv(S).
D
We will show one application of this theorem.
Lemma 3.42
Let K be a compact set. Then Ext(Bc(K)*) — {^k}keK, where 8k are
Dirac Junctionals in C(K)*.
Proof: We will first show that all extreme points of Bc(k)* are of the
form ±5fc. Let A = conv™ {±6k}keK- We claim that A = BC(k)* • Indeed,
assuming F E BC(k)* \ A, by Theorem 3.18 there is / E C{K) such that
supA(/) < 1 and F(f) > 1. Since A is symmetric, supA |/| < 1, which
implies sup \Sk(f)\ = ||/|| < 1- Thus F(f) < 1, a contradiction. Now, by the
keK
tu*-analog of Theorem 3.41, we obtain that Ext(Bc(K)*) C {±6k}k?K
In the proof of Lemma 3.23, we showed that ({6k}keK, w*) is homeomor-
phic to K, so {6k}keK is iu*-compact. Thus also {±6k}keK - {t>k}keK U
{—f*k}keK is tu*-compact, hence u>*-closed; so Ext(BC(K)*) C {i^}^^;.

3. Weak Topologies 79
It remains to show that every Dirac measure 6k is an extreme point of
Bc(K)* • This will complete the proof because then — 6k are extreme by the
symmetry of BC(k)* •
Given &o G K, consider the family U of all open neighborhoods of &o
in K. Given G E W, by Urysohn's lemma there is fu G Bq(k) such that
fu(ko) = 1 and fu = 0 on K \ U. Then Hv = {F e C(K)*; F{fv) = 1}
is a w*-closed supporting manifold for Bc(k)*- Define H — f] Hu-
ueu
Since 6ko G Hu for all U G U, we have that H ^ 0, and it is a
inclosed supporting manifold for Bc(k)*- Since H C\ Bc(k)* is w*-compact,
it contains an extreme point, which is also an extreme point of Bc(k)* (see
the proof of Theorem 3.37). However, the only candidates are then ±6j- for
k G K, and we easily check that 6k ? H for k ^ fco, because the topology
of if is Hausdorff. Thus, 6k0 must be an extreme point.
?
Theorem 3.43 (Banach, Stone)
Let K, L be compact spaces. C{K.) is linearly isometric to C(L) if and only
if K and L are homeomorphic.
Proof: Let <p be a homeomorphism of K onto L. Define a map T: C(K) —»
C(L) by T(f) — f o <p~x. It is standard to check that T is an isometry of
C(K) onto C(L).
Assume now that T is an isometry of C(K) onto C(L). Then T* is an
isometry (Exercise 2.41). Thus T{Bc(k)*) — ^c(L)* > and by Exercise 3.70,
for / G L we have T*Fj) = e(lNkl, where A/ G K and e(/) = ±1. The
mapping g:l *-> e(l)ki is continuous since T* is w*-w*-continuous.
We now show that the function e(l) is continuous. Indeed, for x = 1
we can write e(/) = e(lNkl(x) = T*Fi)(x) = T(ar)(/), so e = T(a?) is
continuous. Then the map p: L —> K defined by /?(/) = kj, p — ^, is a
continuous one-to-one map from L onto J\.
?
Note that C(K) is isomorphic to C(L) whenever K and L are metrizable
uncountable compacts (Milyutin; see, e.g., [Woj]).
Definition 3.44
Let (X,\\ ¦ ||) be a Banach space. Let C be a bounded subset of X*. A set
B C C is called a James boundary for C if for every x G X there is g G B
such that g(x) = sup{/(#); / G C}.
A set B C Bx* is called a James boundary of X if it is a James boundary
for Bx*.
Note that the condition is homogeneous in x. For instance, B C Bx*
is a James boundary of X if for every x G Sx there is / G B such that

80 3. Weak Topologies
Fact 3.45
Let X be a Banach space. Then Ext(Bx*) is a James boundary of X.
Proof: Given x ? Sx, consider H = {/ ? X*; f(x) = 1}. Then H is a
u>*-closed supporting manifold for Bx*, which contains an extreme point
of Bx* by Lemma 3.38.
D
Variants of the following result were obtained independently by many
authors using different methods (e.g., [God2], [God3], [Rod]). Here, we will
follow Godefroy's approach using Simons's inequality (see also [God5]).
Theorem 3.46 (Godefroy [God2])
Let X be a Banach space, and let C be a bounded closed convex subset of
X*. If B is a separable James boundary for C, then C = coirv(B).
In particular, if a separable subset B of Bx* is such that convE) ^ Bx*,
then there is x ? Sx such that x(f) < 1 for all / ? B.
Theorem 3.46 does not hold in general if the assumption on the
separability of B is dropped. For example, consider X = C[0,1] and B =
{±St; t ? [0,1]} C Bx*. Then conv(?) ^ Bx* (Exercise 3.73) and yet, for
every feSx, there is F ? B such that F(f) = 1.
Let B be a nonempty set. Consider the space ^oo(-S) with its canonical
sup-norm. If x — (xb) ? ?OQ(B)) we denote sup^z) = sup{#&; 6 ? B}.
If there is bo ? B such that Xb0 — supB(x), we say that x attains its
supremum over B. In the proof of Theorem 3.46, we will use the following
results.
Lemma 3.47 (Simons's inequality [Sim])
Let B be a nonempty set. Let {xn} be a bounded sequence in l^B). Assume
oo oo
that for all An > 0; n ? N; satisfying ^ An = 1, the vector Yl &nXn
n—l n=l
attains its supremum over B. Then
sup(limsup(?n)) > inf{sup(#); x ? conv{#n}}.
B n B
f OO OO -j
Proof: (Oja [Ojal]) Set Ck = { ? Knxn; A„ > 0, ? An = lj for fc e N.
We must prove that
inf sup(z) < sup(limsup(xn)). (*)
sGC'i Q B n
Let e > 0. Choose inductively Zk ? Ck such that for k = 0,1,...,
supB*Vfc + 2*+i) < inf supBfci;jb + z) +
k z °° z
where ^0 = 0 and vy. — Y2 -^ for k ? N. Put v — Y2 7^~-
n=l^ n=l^

3. Weak Topologies 81
Since zk+1 = 2k+1vk+1 - 2k+1vk, we get 2k+1vk+1 - 2kvk = 2kvk + zk+1,
oo 2,
so using 2kv - 2kVk — 2k ]T ^ ^ Ck+i we get for all fc = 0,1,...,
n=k+l *
6
supBk+1vk+1-2kvk) < supBkvk + Bkv-2kvk)) +
2k+1
ki
B ^'VTA B
Because v E Ci, we can choose t E B such that v(i) = sup5(v). Since
m —1
^ 2* = 2m — 1, from the last inequality we have for m E N:
k=0
m—1
2mvm(t) = ?B*+V+i-2S*)(*)
k=Q
< Bm - 1) sup(t;) + e = 2mv{t) + e - sup(v).
B B
Thus supB(v) < 2mv(t) - 2mvm(t) + e, and hence
inf sup(z) < sup(v) < limsupBm?; - 2mvm)(t) + e
x?Ci ? B m
< limsup(zm(*)) + e
m
using 2mv — 2mvm E Cm+i. Inequality (*) follows because e > 0 was
arbitrary.
?
If X is a Banach space and B C I*, we can consider X a subset of
?oo(B) with ? = (/(#)) fPD. Note that if jB is a James boundary of X, then
sup^z) = ||x|| and the supremum is attained. Thus we get the following
theorem.
Theorem 3.48 (Simons [Sim])
Let B be a James boundary of a Banach space X. If {xn} is a bounded
sequence in X, then
sup{limsup(/(zn)); / E B} = sup{limsup(/(zn)); /E#x*}.
Proof: Assume sup(limsup(#n)) < a < sup(limsup(zn)). Let / E Bx*
B Bx*
be such that limsup(/(a:n)) > a. Assume without loss of generality that
/(#n) > & for all n E N. If x E conv{zn}, from the standard convexity
argument we have f(x) > a. On the other hand, since sup^(x) = ||#||,
from Simons's inequality there is x E conv{#n} such that \\x\\ < a and
thus f(x) < a, a contradiction.
?
We are now ready to prove the Godefroy separation theorem.

82 3. Weak Topologies
Proof of Theorem 3.46: ([God2]) By contradiction, assume that C ^
conv(J9). By the separation theorem, we can find F G Sx** •> & < /?> and
y*Q G C \ convE) such that F(f) < a for all / G ? and F(y%) > /?.
Let S = {x G Bx '• yoC^) > P}- Using Goldstine's theorem, we get that
w*
F G S . Because B is separable, the topology on bounded sets in X** of
pointwise convergence on B is metrizable. Thus, there is a sequence {xn}
in S that converges to F on points of B.
In particular, f(xn) —*¦ F(f) for / G ?, so supB(limsup(xn)) =
sup5(F) < a. On the other hand, conv{zn} C S and yj ? C, so
supjB(x) = s\ipc(x) > /3 for a: G conv{#n}. By Simons's inequality,
a > sup(limsup(?n)) > inf sup(#) > /?,
5 r6conv{a:n.} b
a contradiction.
?
Corollary 3.49 (Godefroy)
Let X be a Banach space X. If X has a separable James boundary, then
X* is separable.
Proof: If B is a separable James boundary of X) then by Theorem 3.46
conv(jB) = Bx* • Thus, Bx* and also X* are separable.
?
By Fact 3.45 we have the following result, which was obtained
independently by several authors (for references, see, e.g., [LoPe]).
Corollary 3.50
Let X be a separable Banach space X. IfExt(Bx*) is separable, then X*
is separable.
We also see how James boundaries interact with second duals.
Corollary 3.51
Let B be a James boundary of a Banach space X. Ifx** G X** is a w*-limit
of a bounded sequence in X, then \\x**\\ — sup (#**(/)).
feB
Proof: Let {xn} be abounded sequence in X such that xn —> x**. Assume
sup{a?**(/); / G B} < |fz**||. Then we can find x% G Bx* and real numbers
a < /? such that, for every / G B,
We may assume /? < xn(xQ) for all n G N. Since B is a James boundary,
by Lemma 3.47 we get a contradiction:
a > sup{***(/); / G B} > _in_f sup{/(z); f EB}>C,
07 6conv{a:n}
?

3. Weak Topologies 83
In particular, we obtain the following corollary.
Corollary 3.52
Let X be a Banach space. If every u E Sx** is the w*-limit of a sequence
in Bx, then Bx — convE) for every James boundary B of X.
Consider a functional / G X*. We say that it attains its supremum over
C if there is x G C such that /(c) = sup{/(#); c G C}. We say that / G X*
attains its norm if there is b G B such that /(&) = ||/||.
Theorem 3.53 (Ekeland variational principle; see, e.g., [DGZ3])
Let (p be a lower semicontinuous bounded-below function from a Banach
space X into RU {+00}. For every e > 0, there is xq G X such that
(p(x) > (p(xo) — e\\x — xq\\ for every x G X.
Proof: Put y\ G X arbitrary and define inductively yn+i G X so that
(p{yn+i) + e\\yn -yn+i|| < M{<p(x) + s\\yn - x\\\ x EX} + ±. (*)
Then (p(yn+i) < ?>(j/n+i) + e\\yn - 2/n+ill < <p(yn) + ^r- Thus, \im(<p(yn))
exists and is finite. Moreover,
?||yn ~ 2/n+l|| < ^(yn) " ?>B/n+l) + ^T-
Thus {||r/n — 2/n+i||} is Cauchy and therefore {yn} converges. Let x0 —
lim(yn). By (*), we have
VKl/n+l) < H>{*) + ^lk ~ 2/n|| + ^r - e||j/n ~ y« + l||
for every xEl and every n G N. By taking the limit for n —> oo,
D
Theorem 3.54 (Bishop, Phelps; see, e.g., [DGZ3])
Let X be a Banach space. The sei of all functionals in X* that attain their
norm is dense in X*.
Proof: Given g G Sx* and e G @, \)} put (p(x) - ||x||2 - g(x) for x EX.
Note that cp(x) > \\x\\2 - \\x\\ > -\ (minimize f(p) — p2 —p on [0, oo)). By
Theorem 3.53, there exists xq E X such that <p(x) > ip(xo) — e^x — ?o|| for
every x EX. Thus,
INI2 - 9{?) > lko||2 - g(x0) - e\\x - x0\\
for every x E X. By the separation theorem used for the epigraph of the
functions on the left-hand side and the subgraph of the function on the
right-hand side of this inequality (both are convex sets), it follows that
there is a continuous affine function h on X such that
INI2 ~ g(x) > h(x) > \\xQ\\2 - g(x0) - e\\x - xQ\\
for every x E X.

84 3. Weak Topologies
Note that h(xo) = \\xo\\2 — g(xo) and the latter part in the previous
inequality yields h{x) — Ii(xq) > —s\\x — xo\\. We claim that xq ^ 0.
Indeed, assume that xq = 0. Then h(xo) — ||0||2 — g@) = 0 and h(x) >
—e\\x\\ for every x G X. Hence, h is a linear functional and \\h\\ < e.
Since ||^|| = 1 and e < |, we get g + h ^ 0. Choose any y G Sx such
that (g + h)(y) — a > 0. The first part of the previous inequality gives
\\x\\2 > (9 + h)(x) f°r aU ^ G ^. Applying this to x — ty for ^ > 0, we get
that t2 > at] that is, t > a for alH > 0. But that contradicts a > 0, so
If ||z|| = ||xo||, then (g + A)(x) < ||*||2 = ||x0||2 = (g + h)(x0). Thus,
(g + h) restricted to{*GX; ||^|| = ||a?o||} attains its supremum at xq and
so does the linear functional / = g -f h — h@). Hence, / attains its norm
and ||/ — #|| < ?, which follows from the fact that h is affine, so
h(x) — h@) = h(x + xq) — h(xo) > — e\\x\\.
D
Theorem 3.55 (James [Jam2])
Let C be a closed convex subset of a Banach space X. C is w-compact if
and only if every f G X* attains its supremum over C at some point of C.
Note that if every / G X* attains its supremum over C, then C is
bounded by Theorem 3.15.
Proof: Assume that C is w-compact. Since every / G X* is ^-continuous,
the image f(C) of the w-compact set C must be compact in R and hence
closed, so the supremum is attained.
We present the proof of the other direction only for a separable space
X. By our assumption, C is a separable James boundary for C in X**;
hence, by Theorem 3.46, conv(C) = C . Since C is closed and convex, we
w*
get C — C , and hence C is w-compact.
?
The proof in the nonseparable case is much more difficult (see, e.g.,
[Dis2]).
Corollary 3.56 (James [Jam2])
A Banach space X is reflexive if and only if every f G X* attains its norm.
We will now show another of James's characterizations of reflexivity.
Theorem 3.57 (James [Jami])
A Banach space X is reflexive if and only if there exists 0 G @,1) such that
whenever {xn} is a sequence in Sx with inf{||u||; u G conv{xn}} > 6, then
there are n0 G N, u G conv{#i,... ,xno}, and v G conv{zno+lj xno+2, ¦ ¦ •}
such that \\u — v\\ < 0.

3. Weak Topologies 85
Proof: (Oja [Oja2]) Assume that X is reflexive. Fix 0 E @,1), and let {xn}
be a sequence in Sx- For an integer n > 0, let Kn — conv{#n+i, xn+2, • • •}•
Since {Kn} is a nested sequence of nonempty weakly compact sets, there
is x E f]Kn. Because x E Ko> there is no and u E conv{#i,..., xno} such
that ||z-u|| < 0/2. Because x E Kno, there is v E conv{zno+i, xno+2, • • •, }
such that \\x - v|| < 0/2. Then \\v - u\\ < 0.
Let X / X** and any 6 E @,1) be given. We will construct a sequence
{xn} failing the above condition. Denote Be = {F 6 ***; ||-F|I < e)- BY
Lemma 1.23, there is Fe E Sx** \ \J (x + Be).
Choose xo E Sx • There is a convex tu*-neighborhood V\ of Fe that is
disjoint with xq + Be. Thus \\v — a?o|| > # for all v E V\.
Choose x\ E Vi f) Sx- Because Fe ? (conv{#o>?i} + Be)> there is a
convex ^-neighborhood V2 C V\ of Fe such that \\v — u\\ > 9 for all
u E conv{a?o, x{\ and all t; E V2.
Choose a?2 E V^flSx and continue by induction. The sequences of convex
sets Vi D V2 D ... and elements {xn} C Sx satisfy xn E Vn, and also
\\v — u\\ > 6 for all u E conv{iCi,..., xn} and all t; E V^+i, in particular for
v E conv{a?n+i, xn+2,...}. Thus, the condition fails.
?
We will now show several applications of Theorem 3.55.
Theorem 3.58 (Krein)
Let X be a Banach space. If C is a weakly compact set in X, then conv(C)
is weakly compact in X.
Proof: If / E X*, then sup(/) = sup (/). Since C is weakly compact,
C conv(C)
there is x E C such that supc(/) = f(x). Therefore, each / E X* attains
its supremum over conv(C), and due to Theorem 3.55, conv(C) is weakly
compact.
?
The following theorem is a special case of a general result presented
in Chapter 4. Recall that a topological space (T, r) is called sequentially
compact if every sequence in T has a subsequence convergent in (T, r). A
subset C of a Banach space X is called weakly sequentially compact if it is
sequentially compact in its relative weak topology.
Theorem 3.59 (Eberlein, Smulian)
Let C be a weakly closed subset of a separable Banach space X. C is weakly
compact if and only if C is weakly sequentially compact.
Note that if C is weakly (sequentially) compact, then f(C) is bounded
for every / E X* by the weak continuity of /, and therefore C is bounded
by Theorem 3.15.

86 3. Weak Topologies
Proof: If C is ^-compact, then by Proposition 3.29 it is metrizable in the
w-topology, and hence C is ^-sequentially compact.
Assume now that C is weakly sequentially compact, and let D —
conv(C). Given / ? X*, let cn ? C be such that lim(/(cn)) = supc(/).
By our assumption, there is a subsequence cnk of cn such that cnk —> c
for some c ? C, so /(c) = supc(/) = su-pD(f). Hence /(c) — sup^(/).
Thus, / ? X* attains its maximum on D, and D is weakly compact by
Theorem 3.55. Since C is a weakly closed subset of D, it is weakly compact
as well.
?
Note the similarity of the following result with the Lebesgue dominated
convergence theorem for continuous functions.
Theorem 3.60 (Rainwater [Rai], Simons [Sim])
Let B be a James boundary of a Banach space X. Let {xn} be a bounded
sequence in X and x ? X. If f(xn) —> f(x) = 0 for all f ? B, then xn -^> x.
Proof: Since sup(limsup(#n — x)) = sup(limsup(#n — x)) = 0 (using
Bx* B
Theorem 3.48 for the sequence {xn — #}), we get limsup(/(:rn — x)) < 0
for / ? Bx*- Similarly, by considering the sequence {x — xn}, we get
limsup(/(# — xn)) < 0 for / ? Bx*- Hence f(xn) —> f(x) = 0 on Bx*,
and so a?n -> x.
?
Corollary 3.61 (Rainwater [Rai])
Let X be a Banach space, and let {xn} be a bounded sequence in X and
x eX. If f(xn) -> f(x) for every f ? Ext (By), then xn ^ x.
We refer to [FLP] for an approach to James boundaries via the Choquet
representation theory (c/, Theorem 4.43).
Exercises
3.1 Let X be a real Banach space, let f,g ? Sx*, and let e > 0 be such
that \f(x)\ < e for every x ? ?-1@) H Bx. Prove that either ||/ - g\\ < 2e
or ||/ + ^|| < 26:.
Hint: Consider / on <7-1@) and extend it with the same norm (at most e) on
X, calling this extension /. Then / — / = 0 on 5f~1@) and thus f — f — ag
for some a by Lemma 3.9. Note that |l-|a|| = |||/|| - ||/- /||| < ||/|| < e.
Thus, if a > 0, then \\g + f\\ = ||A - a)g + /|| < |1 - a\ + ||/|| < 2e. If
a < 0, calculate ||# — /||.

3. Weak Topologies 87
3.2 Let {xn} be a sequence in a Banach space X. Prove that if Xfi * X j
then {xn} is bounded and \\x\\ < liminf \\xn\\.
n
Let {/n} be a sequence in X*. Prove that if fn —» /, then {/n} is bounded
and ll/H <liininf||/n||.
Hint: Proof is similar to that of Corollary 3.13.
3.3 Let {xn} be a sequence in a Banach space X. Prove that xn —» x if
and only if {a?n} is bounded and the set {/ G X*] f(xn) —> /(a:)} is dense
inX*.
Similarly, let {/n} be a sequence in X*. Prove that fn —> / if and only
if {/n} is bounded and the set {# G X; fn(%) —* /(#)} is dense in X.
In particular, we have the following corollary. Assume that {xn} is
bounded and M C X* is such that span(M) = X* and f(xn) —> /(#) for all
f ? M. Then #n —> #. An analogous statement is true for it;*-convergence.
Hint: One direction was done in the previous exercise and in the chapter.
The other one is standard: let / G X*. Given e > 0, approximate f by g
such that \\g — f\\ < e and g(xn) —> g(x). In the estimate
\f(*n) ~ f(x)\ < \f(xn) - g(xn)\ + \g(xn) - g(x)\ + \g(x) - /(x)|,
we pass to the limit and obtain
limsup \f(xn) - f(x)\ < esup ||arn|| + 0 + e\\x\\.
Since e > 0 was arbitrary, we get f(xn) —> f(x) as needed.
3.4 Let {xn} be a sequence in a Banach space X. Assume that f(xn) —> 0
for every / G A C X*, where A is a set of second Baire category in X*.
Show that xn —> 0.
Hint: See the proof of the Banach-Steinhaus theorem.
3.5 Let X be a normed space. Show that if {xn} is Cauchy and xn ^ 0,
then xn —* 0 in norm.
Hint: a?n G xm -f- sBx j and #m -[- sBx is weakly closed.
3.6 Assume that {zn} is a weakly convergent sequence in an infinite-
dimensional Banach space X. Show that conv{#n} does not have any
interior point. This is not the case for the w*-convergence, as the standard
unit vectors of l\ show.
Hint: (Vesely, Zanco) Assume that xn —> 0 and that K = conv{xn}
contains an interior point 0. Let x be an arbitrary point of the topological
boundary of K. Using the Minkowski functional of K, we construct a
functional / supporting K at x. Let N be such that f(xn) < f(x)/2 for n > N.
Put A — conv{zn; n < N} and B — conv{xn; n > N}. Then there are
Xk G [0,1], ak G A a bk G 5 such that the points j/fc = A — Afc)aA; + Xk^k

88 3. Weak Topologies
tend to x. By applying /, we can see that A& —> 0 because otherwise /
could not have a maximum at x. Hence aj~ —» x, so A is closed. Thus
x € A C conv{zn}. Hence, X would have a countable Hamel basis, a
contradiction.
3.7 The Josefson-Nissenzweig theorem asserts that, for every Banach space
X, there is a sequence {/n} C Sx* such that fn —» 0 ([Dis2]).
Show that, given / E 5j*, there is a sequence {/n} C Sx* such that
Hint: Consider the sphere S centered at / with radius 1. Then there are
w*
gn E S such that (gn — /) —> 0. Project all gn onto Sx in & radial way;
namely, given gn, let fn be the intersection of Sx and the ray emanating
from / and going through gn. Then fn = / + cn(#n — /) for some cn > 0
(use the fact that / E -Bx*, draw a picture). We also have cn < 2 and hence
(/n — /) —> 0 as needed.
3.8 Show that if X is infinite-dimensional, then 0 E Sx .
Using this, as in the previous exercise we can show that Sx = Bx for
infinite-dimensional spaces.
Hint: Consider a neighborhood V = {x;|/{(x)| < e for i = l,...,n}.
Since all fr\0) are 1-codimensional (Exercise 2.5), p|/t- x@) is finite-
codimensional. Because X is infinite-dimensional, there is x E Sx H
3.9 Let {e2} be an orthonormal basis of a Hilbert space H. Show that
e,- —> 0.
Hint: (e2-, e^) —> 0 as f —> oo for each j, and span{e7} = H = if*.
3.10 Show that in the space -?i, 0 is not in conv{e;}.
Hint: Consider A,1,1,...) E ^oo-
3.11 Let X — l\ = Cq. Show that the standard unit vectors et- satisfy
e,- —* 0 but not et- —> 0.
Hint: Show that e2(ar) —> 0 for x E Co.
For the second part, use the previous exercise.
3.12 Show that in the space ^oo5 the set {ei} U {0} is weakly compact but
not norm compact.
Hint: The set is weakly compact in cq since ei —> 0 in cq.
3.13 Show that the set {e,-} of the standard unit vectors in ?2 is norm
closed but not w-closed in ?2-
Hint: If xn E {es-} and xn —> x in I2, then {xn} is eventually constant.
Recall (Exercise 3.9) that the origin is in the ^-closure of {ei}.

3. Weak Topologies 89
3.14 Let {ei}f2zl be the standard unit vectors in ?2. Show that A =
{ei}fl1 U {0} is a weakly compact set in ?2.
Hint: By Exercise 3.9, e; —» 0.
3.15 Let {e2} be the sequence of the standard unit vectors in ?2. Show
w
that 0 G {\/™en} and that no subsequence of {y/nen} converges weakly
to 0. Note that this fact also shows that the weak topology of ?2 is not
metrizable.
Hint: Let V be the neighborhood of 0 given by vectors x1, x2,..., xn in ?2
n
and e > 0. Consider the element z G ?2 defined by Z{ = Y2 \xi\- Note that
*=i
for an infinite number of indexes i we have \z{\2 < | since otherwise z ? ?2.
Therefore, for an infinite number of indexes z, we have \y/iei(xk)\ < e for
k — 1,..., n, in particular V D {V7e2} ^ 0. The second part follows from
Exercise 3.2.
3.16 Let fn G C[0,1], ||/„|| < 1. Show that fn ^ 0 in C[0,1] if and only if
fn(x) —> 0 for every x G [0,1].
This shows that for bounded sequences in C[0,1], weak convergence is
equivalent to pointwise convergence. An analogous statement is true for cq.
Hint: Use Corollary 3.61 and Lemma 3.42.
3.17 Let x,xn G ?2 be such that xn —» x in ?2. Show that there is a
subsequence {xnje} such that the Cesaro means
%n\ "T #n2 -f- . . . -r xrik
k
converge to x in ?2 (the Banach-Saks theorem).
Show that an analogous statement is true for Co.
Spaces that have this property are called spaces with the weak Banach-
Saks property.
Hint: The ?2 case: assume x = 0. Let ||a;n|| < M for all n. Put ri\ = 1 and
if rii,..., rik were chosen, pick rik+i such that |(zni, #nfc+1)| < \ for every
i < *. Then
l|gWl + xn2 + ... + gWjb+1 [J2 (* + 1)M2 + 2-l + 2-f + ...+ ^
(*+lJ " (Jb + 1J "^
The Co case: the sliding hump argument.
3.18 Define a functional / on ?\ by f(x) — ^ #; for x — (xi). Show that /
is norm continuous on ?\ and that / is not w*-continuous on ?\ ¦= Cq (that
is, / G4o \c0).
Hint: Consider the standard basis {en} of ^1. Note that en ^-> 0 in ^1.

90 3. Weak Topologies
3.19 Let X and Y be Banach spaces. Show that if a linear operator T from
X into Y is w-w-continuous, then T ? B(X,Y). On the other hand, every
bounded linear operator is w-w-continuous.
Hint: Given / E Y*, consider V — f~l{—1,1)- Then V is a ^-neighborhood
of zero in Y. By our assumption, T_1(V) is a ^-neighborhood of zero in
X. Therefore, XBX C T^) for some A; that is, \f(T(Bx))\ < j. Thus
T(BX) is ^-bounded and also bounded in Y. On the other hand, if xa —» 0
in land/GY*, then / o T ? X* and thus T(xa) -^ 0 in Y.
3.20 Let T be a bounded linear operator from Y* into X*. Show that T
is a dual operator if and only if T is w*-w*-continuous.
Hint: Assume T is w*-w*-continuous. Given x E X, the function 7r(x)oT is
w*-continuous on Y*, so ir(x)oT = x(T) E Y. The assignment x »-» t(x)oT
defines the predual operator. The other implication is straightforward.
3.21 Let X be a separable Banach space, and let {xn} be a dense sequence
in Sx. Define a mapping T from X* into ?2 by T(f) = (f(xi)/2i). Show
that T is a bounded linear operator that is continuous from the tu*-topology
of X* into the ^-topology of ?2.
Hint: To check that T is linear and ||T(e)||2 < \\x\\ is routine. Note that
T*(e*) = §*¦, so by continuity of T* and reflexivity of ?2 we get T*:?2-> X.
Thus T = T**, which is u>*-i/;*-continuous as a dual operator. Since ?2 is
reflexive, the claim follows.
3.22 Let X,Y be Banach spaces and T ? B(X,Y). Show that:
(i) T is an isomorphism into if and only if T** is an isomorphism into,
(ii) T is an isometry into if and only if T** is an isometry into.
Hint: (i): If T is an isomorphism into, it follows directly from Exercise 2.39
using (i) and then (ii) applied to T*. If T** is an isomorphism into, we have
||T**(?**)||y** > <$||?**||x** for some 6; hence, the same holds for T, and
then use Exercise 1.27.
(ii): Going through the proof of Exercise 2.39, we see that T*{By*) =
Bx*> which implies ||T**(F)||y~ = ||F||x--
3.23 Let X}Y be Banach spaces, T E B(X,Y). Show that if T is an
isomorphism of X onto Y, then T**(X) = Y and (T**)^) = X.
Hint: T is onto Y, hence T**(X) = T{X) = Y. Also, T** is an isomorphism;
in particular, T** is one-to-one, so (T**)-1(Y) C X.
3.24 Let X, Y be Banach spaces and T E B(X, Y). If T: X -+ Y is one-to
one, is T** necessarily one-to-one?
Hint: No. Consider the identity map from ?\ into ?2. Its dual operator
cannot be onto a dense set in i^ because ^00 is not separable. Since T* is
not onto a dense set, T** is not one-to-one by Exercise 2.36.

3. Weak Topologies 91
3.25 Let X be a separable Banach space. Find T G B^^X) such that
Til?) is dense in X. Note that then T*:X* —» ?2 is a bounded linear
iu*-w-continuous one-to-one operator.
Hint: Let {yi} be dense in Bx and T be defined for a? = (xi) G ^2 by
T(x) = Yf2-ixiyi. show that 2/f € TD0 for every i.
3.26 Show that there exists a sequence {/n}?°=i in the dual of some normed
linear space X such that {/n(^)}^°=i is bounded for every x ? X and yet
{||/n||} is unbounded. This shows that the assumption of completeness of
X cannot be dropped in the Banach-Steinhaus theorem.
Hint: In Cqo, put fn = nen, where en is the standard unit vector. Then
check that for every x G cqq we have fn(x) = 0 for n large enough.
3.27 Show that every weakly compact set C in a Banach space is a bounded
set.
Hint: If / G X*, then f(C) is compact in K since every / G X* is w-
continuous. Then use the Banach-Steinhaus theorem.
3.28 Let {fn} C X*. Show that if there exists en —> 0 such that for every
a?GX there is i^ > 0 with |/n(a?)| < Kxen, then ||/n|| -*¦ 0.
Hint: The Baire category.
3.29 Let X be a Banach space. Assume that xn —* x and fn G X* are
such that fn ^ /. Show that fn(xn) -+ /(^)-
Hint: Write fn(xn) - f(x) = /n(a?n) - fn{?) + /n(^) - /(a?) and note that
{/n} is bounded by Banach-Steinhaus.
3.30 Assume that || • \\x and || • ||2 are norms on a normed space X that
are not equivalent. Show that then there is a linear functional on X that
is continuous in one of the norms and is not continuous in the other.
Hint: One of the unit balls in these norms is not a bounded set in the
normed space whose unit ball is the other unit ball. This unbounded ball is
not weakly bounded in the topology given by the other norm. Thus, there
is a functional continuous in the other norm and not in this one.
3.31 As an application of the Banach-Steinhaus uniform boundedness
principle, show that there is a continuous 27r-periodic function whose
Fourier series diverges at 0.
Hint: Let C be the Banach space of 27r-periodic continuous functions on
n
R with the supremum norm. For / G C, put crn{f) = Yl ck> where
k — — n
ck = ^ fln f{t)e~lktdt is the &-th Fourier coefficient of/. Thus crn(f) =
&flwf(t)I>n(t)dt, where Dn(t) = ? e~ikt = sin((n+j)*) (that is>

92 3. Weak Topologies
an(f) is the value at 0 of the n-th symmetric partial sum of the Fourier
series of /). Thus, an is a continuous linear functional on C and ||crn|| =
^F Slv \Dn(t)\ dt. We estimate
dt
sin((n+|)t)
1_ rT\sm((n+\)t)
K JO
- i r
"~ * Jo
Jo
t
sin((n+|)t)
t
sin(k)
dt
dt
Jo
("+*)
sin(u)
c/u
and this expression tends to infinity as n —> oo since J0 l^—^l du diverges.
Thus supn ||crn|| = oo. By the Banach-Steinhaus theorem, there is / G C
such that {crn(f)}^L1 is unbounded, so the Fourier series for / diverges at
0.
w*
3.32 Find a Banach space X and fn G X* such that fn —> 0 and yet for
every convex combination h of {/n} we have \\h\\ = 1.
This shows that Mazur's theorem does not hold for the if;*-topology.
Hint: Put X = c0, /n = @,..., 0,1, 0,...) G h.
3.33 Let C be a bounded set in a Banach space X. Show that C is
incompact if and only if C C X.
w*
Hint: C is tu*-compact (Alaoglu).
3.34 Let X be a Banach space. Assume that C is a tc*-compact set in X*.
Is conv™ (C) also u>*-compact?
Hint: Yes, because conv^ (C) is a iu*-closed subset of some dual ball, which
is w*-compact by Alaoglu's theorem.
3.35 Can we replace Bx and Bx** in Goldstine's theorem by Sx and
Hint: Yes. Use the iu*-lower semicontinuity of the second dual norm,
w*
namely, if xa G X and xa —> x** in X** for some x** G X**, then
limsup ||^or|| > lk**||-
3.36 Prove the following Helly's theorem: Let X be a Banach space. If
x** G Sx**, F is a finite set in X* and e > 0, then there is x G X with
||z|| < A +e) such that /(a?) = x**(f) for every / G F.
Note the difference between Helly's and Goldstine's theorems.
Hint: Put Z = FL. Consider the space X/Z and x*Q* G (X/Z)** = (X/Z),
since X/Z is finite-dimensional. Then find in the coset xj* an element
x G X of almost the same norm as ?q* .

3. Weak Topologies 93
Note that e cannot in general be taken to be 0. Let / G Sx* be such
that it does not attain its norm, and take x** G Sx** such that #**(/) = 1.
Then there does not exist any x G Bx such that f(x) = x**(f) = 1.
3.37 Let X be a Banach space. In Exercise 2.21, we proved that for a
subset Y of X we have (YL)±_ = span(Y). Since span(Y) is convex, by
Mazur's theorem it also reads (YL)± = span^(y).
Prove that for a subset Y of X* we have (Y±)L = span*" (Y).
For T G B(X,Y), we then have T(X)W = Ker(T*)± and T*(Y*f* =
Ker(T)x.
Hint: Show that (YjJ1- is w*-closed, so span*" (Y) C (XlI- Assuming that
the inclusion is strict, derive a contradiction using Theorem 3.18.
3.38 Let X,Y be Banach spaces and T G B(X,Y). Consider f as an
operator from Xj Ker(T) into T(X). Show that then T** can be considered
an operator from X**/ Ker(T**) into T(X)**; in particular, f**(x) = T(s)
for all x G X (see Exercise 2.35).
Show also that under the isometry of (X/ Ker(T))** onto X**/ Ker(T**),
the subspace Xj Ker(T) of {Xj Ker(T))** corresponds to the subspace {x-\-
Ker(T**); x G X} of X**/Ker(T**).
Hint: We have T*:T(X)* -+ (X/Ker(T))*, so f**: (X/Ker(T))** -+
T(X)**.But
(X/Ker(T))** = (Ker(TI)* = (T*(X*)"*)* = X**/T*(X*;f *\
But in general, IT = Z1 for any subspace Z of X*, so (X/ Ker(T))** =
X**/T*{X*)L = X**/Ker(r**).
3.39 Let X,Y be Banach spaces, T G B(X,Y). Assume that T(X) is
closed, and let x** G X**. If T**(z**) G T(X), then for every 8 > 0
there exists xeX such that T{x) = jT**(;z**) and ||z|| < A + <$)||z**||.
Hint: Assume that T{X) = Y; then T:X/Ker(T) —> Y is an isomorphism
and thus f**:X**/Ker(T**) -» Y** is an isomorphism (Exercises 3.38
and 2.41). Since f**(z**) G Y, we have x** = x + Ker(T**) for some
x E X. Thus, x** corresponds to some x G X/KerCT) (Exercise 3.38), and
hence there is x G x such that yzj||#|| < \\x\\ = ||#**|| < ||e**||-
Let X be a Banach space and A C X*. We say that A is separating
(or separates points of X) if for # G X, x / 0, there is / G A such that
f(x) ^ 0. Equivalently, for any x ^ y in X there is / G ^4 such that
We say that A is c-norming if sup{/(#); / G ^4H^x*} > ^\\x\\ for every
x G X. We say that A is norming if it is c-norming for some c > 1.

94 3. Weak Topologies
By the Hahn-Banach theorem, both X* and Bx* are 1-norming. Also, X
is 1-norming for X*. We will use the term norming only for closed subspaces
ofX*.
3.40 Show that every norming subspace of X* is a separating set for X.
Find an example of a separating subspace that is not norming.
Hint: For the second statement, take a partition 7, Ii,/2>--- °f N into
disjoint infinite sets. Let S = < y G t\\ for each k G 7, y^ = ? ]P yn >.
1 ne/fc j
Then 5 is separating for Co but not norming (consider ej. for k ? I).
Note that if X is separable and X** /X is infinite-dimensional, then there
is a separating subspace of X* that is not norming (see [DaLi]).
3.41 Let X be a Banach space and A C X*. Show that A separates the
points of X if and only if A± = {0} if and only if conv™ (A) = X*.
Hint: Easy from the definition and Exercises 3.37 and 2.20.
3.42 Let X be a separable Banach space. Show that there is a sequence
{/n} C Sx* such that {/n} is separating in X.
Hint: Let {xn} be dense in Sx- For each n, choose /n G Sx* sucn that
fn(xn) = 1. Given # G Sx, take n such that \\x — xn\\ < |; then fn(x) > |.
3.43 Let FGl**\ X. Show that F^) is a norming subspace of X*.
If a? G X \ {0} C X**, can ^(O) C X* be norming?
Hint: Let the distance of F to X be 25 > 0. Assume that for some x G Sx
we have sup{/(x); / G ^"H0); ||/|| < 1} < 5/2. Then, by Exercise 3.1, we
have \\x ± F\\ < 5, a contradiction with the distance of F to X.
If x G X \ {0}, a?_:L@) cannot be norming because it is iu*-closed. If
it were norming, it would be separating, hence iu*-dense in X*, and thus
ar-1@) = X*, a contradiction.
3.44 Let X be a reflexive space, and let Y C X* be a closed subspace of
X*. Show that if Y separates points of X, then Y — X*.
Hint: We have Y™* = X*. By reflexivity, F™ = X*; then use Theorem 3.19.
3.45 Let T be a linear operator from a Banach space X into a Banach
space Y. Suppose that F is a subset of Y* that separates points of Y.
Assume that f(T(xn)) —*¦ 0 whenever f € F and {#n} C X is such that
||^n|| —* 0. Show that T is a bounded operator.
Hint: The closed graph theorem.
3.46 Prove directly that if X is a separable Banach space, then X* is
iu*-separable.
Hint: Take the sequence {/n} C Sx* from Exercise 3.42. Then span{/n} is
tu*-dense. Therefore, the rational combinations form a tz;*-dense set in X*.

3. Weak Topologies 95
3.47 Let X be a Banach space. Show that the following are equivalent:
(i) Bx is ^-separable.
(ii) Sx is inseparable,
(iii) X is separable.
Hint: (i) => (ii): If a countable C C Bx is to-dense in Bx and x ? Sx,
then there is a net {xa} C C such that xa —> x. We can assume that
\\xa\\ —> 1 because the norm is weakly lower semicontinuous, and hence
p?i| ^ x- {]!%'» ^ ^ C} is w-dense in Sx.
Every weak neighborhood of a point from Bx intersects Sx if X is
infinite-dimensional. Thus (ii) => (i). Clearly, (i) implies that X is weakly
separable and thus norm separable (Proposition 3.26). (iii) implies that the
ball and the sphere are separable and hence w-separable.
3.48 Let X be a Banach space. Show that Bx* is w*-separable if and only
if Sx* is ^-separable.
Note that the w*-separability of Bx* is not preserved when passing to
an equivalent norm (Exercise 12.40). Also, there is a space X for which
X* is ^-separable and no equivalent norm on X exists so that Bx* is
to*-separable ([JoLl]).
Hint: Similar to the previous exercise.
3.49 Show that t^ is ^-separable.
Hint: Goldstine's theorem and t\* = t^.
3.50 Show that every weakly compact set in ^oo is norm separable.
Hint: Let C be w-compact in ^oo- ^ ls w*-separable, so C is w-metrizable
by Proposition 3.29, and hence also ^-separable. Thus span(C) is weakly
separable and thus norm separable.
3.51 Let X be an infinite-dimensional Banach space. Show that the w-
topology of X is not first countable; in particular, it is not metrizable. The
same is true for the u;*-topology of X*.
Hint: If {Vn} were a countable basis of neighborhoods at 0, assume without
loss of generality that Vn — {x ? X] max |/f (x)\ < sn}. Since X* has
»"=l,...,iVn
no countable Hamel basis, there is / ? X* such that / is not in the span
of all /f. We claim that U = /_1(—1,1) (which is a weak neighborhood
of 0) does not contain any of these Vn. Indeed, if Vn = {x\ max|/z(x)| <
en} C U, then H/T^O) C /-1@) and / would be a linear combination of
{fi}. The proof for X* is similar.
3.52 Use the following hint to show that the weak topology of an infinite-
dimensional Banach space is never metrizable.
Hint: Assume that the weak topology of a Banach space X is metrizable
by a metric p. We know that 0 is in the weak closure of nSx for every n.

96 3. Weak Topologies
Therefore, we can find a point xn E nSx for which p(xn,Q) < ^. Then
p(xn, 0) —» 0, which means that zn -* 0 in the weak topology. Thus, by the
Banach-Steinhaus theorem, {xn} must be bounded, which is not the case.
3.53 Let Y be a closed subspace of a Banach space X. Show that the weak
topology of X/Y is the factor topology of the weak topology of X; that is,
U is weakly open in X/Y if and only if q~l(U) is weakly open in X, where
q is the canonical quotient map.
Show that if T is a bounded linear operator from a Banach space X onto
a Banach space Y, then T is an open map in the respective weak topologies
ofX and Y.
Hint: (X/Y)* = Y1, and every bounded linear map is w-w-continuous.
3.54 Show that every Banach space X in its weak topology is a completely
regular space; that is, if p ? C, where C is a weakly closed set, then there
is a weakly continuous function / on X such that f(p) = 1 and f(x) — 0
for every x E C.
Hint: If for instance 0 ^ C for a weakly closed C, then there is e > 0 and
a finite number /,- E X* such that for V = {x\ max|/2(x)| < e} we have
V fl C — 0. Consider the function r o max |/t-|, where r is a continuous real
function such that r@) = 1 and r(t) — 0 whenever i > e.
3.55 Let X be an infinite-dimensional Banach space. Show that (X*, w*) is
not a complete topological space (see the definition in Chapter 4). A similar
result holds for the u>-topology. Recall that a sequence {xi} is w-Cauchy if
if(xi)} ^s Cauchy for every / E X*.
Hint: Consider a discontinuous linear functional / on X. For every finite-
dimensional subspace of X with a basis B and e > 0, consider a continuous
linear functional g on X such that / and g differ by at most e on vectors in
B. By ordering B by inclusion, one can get a net of continuous linear
functional converging to /. This net is thus u>*-Cauchy and not w*-convergent
inX*.
3.56 Let X be an infinite-dimensional Banach space. Show that Sx is a
dense G§ set in (Bx,w). Deduce from this that the norm of an infinite-
dimensional space is never a weakly continuous function on X.
Hint: Given \\x0\\ < 1, let U — {x\ \ft(x — xq)\ < e}. Take x ^ 0 such that
x E fl/r1@) and n°te that the continuous function || • || considered on the
set {xo+tx; t E R} attains the value 1 somewhere, showing that UdSx ^
0. Therefore, Sx is w-dense in Bx- Put Gn = {x E Bx] \\x\\ > 1 — ^}.
Then the sets Gn are open in (Bx,w) and Sx — f)Gn.
3.57 Let X be a Banach space and || • || be the dual canonical norm of X*.
Show that II • II is a u>*-lower semicontinuous function on X*.
Hint: Alaoglu's theorem.

3. Weak Topologies 97
3.58 Construct an equivalent norm on i\ ~ c? that is not u>*-lower
semicontinuous.
Hint: Consider the norm ||(#j)|| = ]?|#»| + 2|^?2-|. Then take a2- < 0
such that Y^, ai — ~ 1 an(i consider the vectors xn = A, 0,..., 0, ai, c*2,...),
w*
where a\ is at the n-th place. Show that xn —¦» A,0,...) and compare the
norms.
3.59 Let X be an infinite-dimensional Banach space. Can X be a Baire
space in its weak topology? Recall that a Baire space means that an
intersection of open dense sets is dense.
Hint: No. nBx are nowhere dense in (X, w).
3.60 Show that 0 is not a G§ point of ^2(r) m the weak topology if T is
uncountable. Recall that p ? W is a Gs point of W if {p} = p|On, where
On are open in W.
Hint: If {p} = f]On and On are basic neighborhoods in the weak
topology, then each On restricts only a countable number of coordinates, so
p| On restrict only a countable number of coordinates and thus cannot be
a singleton if T is uncountable.
3.61 Let X be a reflexive Banach space. Show that if Y is isomorphic to
X, then Y is reflexive.
Hint: By is ^-compact as a w-closed (see Mazur) subset of T(||T_1[|.?x),
which is ^-compact by the w-w-continuity of T.
3.62 Show that cq is not isomorphic to a subspace of ?p, p G A, co).
Hint: Co is not reflexive.
3.63 Let Y be a closed subspace of a reflexive Banach space X. Show that
X/Y is reflexive.
Hint: (X/Y)* = YL C X* and X* is reflexive.
3.64 Let Y be a closed subspace of a Banach space X. Show that if Y and
X/Y are reflexive, then X is reflexive. Thus, reflexivity is a three-space
property.
Hint: If Y is reflexive, then Y is uAclosed in X**. Thus Y = (y1I. Then
X**/YL± = (y1)* and y* = (X/Y)*, so (y1)* = (X/Y)** = X/Y.
Consequently, X — X**.
Note that this is a delicate proof although it looks easy. We must prove
not just X — X** but actually n(X) — X**, that is, we must check that
the maps in these equalities work as they are supposed to.
3.65 Let X be a reflexive Banach space, and let T be a bounded linear
operator from X onto a Banach space Y. Show that Y is reflexive.

98 3. Weak Topologies
Hint: Y is isomorphic to X/Ker(T), which is a reflexive Banach space.
3.66 Show that if T is a one-to-one operator from t\ into t^ then T{?\) is
not closed in ?2.
Hint: Otherwise, T would be an isomorphism of ?\ onto a reflexive space
T(?i), and hence ?\ would be reflexive.
3.67 Let X be a Banach space. Show that:
(i) If the iD-topology and the uAtopology coincide on X*, then X is
reflexive.
(ii) If the w-topology and the iu*-topology coincide on X**, then X is
reflexive.
Hint: (i): Use Theorem 3.16. (ii): Use Alaoglu's theorem and Theorem 3.31.
3.68 Let X, Y be Banach spaces, T G B{X, Y). Show that if X is reflexive,
then T(BX) is closed in Y.
Hint: Bx is weakly compact and T is weakly continuous.
3.69 Assume that the unit sphere of a given Banach space X contains an
infinite sequence of points such that the mutual distance of two different
members of the sequence is 2. Does it imply that X is nonreflexive?
Hint: No. Consider the equivalent norm on ?2 defined for x — {x{} by
\\x\\ = max{||a?||2,max{|a?i| + |a:j|; i ? j}}•
Check the standard unit vectors.
3.70 Let X, Y be Banach spaces.
(i) Let T be a one-to-one map from X onto Y. Show that if C is a convex
subset of X, then T(Ext(C)) = Ext(T(C)).
In particular, if T is an isometry of X into Y, then T(Ext(Bx)) =
Ext(T(Bx)).
(ii) Let T be a bounded linear operator from X into Y. Show that if C is
^-compact in X, then Ext(T(C)) C T(Ext(C)).
Hint: (i): If x G C\ Ext(C), then x = |(t/ + z) for some y,z G C. But then
also T(x) — \(T{y) + T(z)). Since T\ is invertible, the other direction
follows.
(ii): Let d G Ext(T(C)). Denote K = C n T~1(d). Since T is w-w-
continuous, K is ^-compact and hence there is c G Ext (if) such that
T(c) = d. Check that c G Ext(C): if c = |(ar + y), x,y E C, then d =
|(T(ar) + T(y)). Since d is extreme, T{x) = T(y) = T(d), so x}y G if.
Since c is extreme in if, x — y.
3.71 Let C be a convex set in a Banach space X, and let c be an extreme
n
point in C. Show that if c = ]T A^ with ^ Az- = 1, A2- > 0, and X{ G C
for f = 1,..., n, then c = a:,- for some z.

3. Weak Topologies 99
Hint: A^i + \2x2 + A3z3 = (Ai + A2)(x^^-xi -f- j^^x2) + X3x3} use
induction on n.
3.72 Show directly that Bloo = conv(Ext(B/eo)).
Hint: Note that (xn) is extreme in Bi^ iff \xn\ — 1 for each n. Given
x G Bt^ and e > 0, find a simple function y on N such that \\x — y\\ < e.
If {a,i}i=1 are the values of y) write {a,i}f=1 as a convex combination of
extreme points in B^ . Then express y as the same combination of the
characteristic functions corresponding to {a2}f, which are extreme points
in Bioo.
3.73 Show that conv(Ext(jBc[o,i]*)) / -#c[o,i]*-
Hint: Show that the Lebesgue measure A is not in conv(Ext(i?c[o,i]*))-
Indeed, if A is norm-close to Ylaihn ^n(^ / ^ ^[o,i] such that f{ti) = 0
but A(/) is close to 1.
3.74 Show that BCo has no extreme point.
Hint: If x = (xi) G BCo, let z'o be such that \x{0\ < \. Then a? is the center
of a line segment with endpoints (x^ ± |ez-0.
3.75 Show that extreme points of Bc are vectors x = (±1).
Show that c and Co are not isometric.
Hint: Isometry preserves extreme points; use the previous exercise.
3.76 Show that the extreme points of Bc[o,i] are exactly two functions:
constant functions 1 and —1.
Show that Bc(k) has some extreme points for arbitrary compact set K.
Hint: Use the definition.
3.77 Show that Co is not isometric to any C{K) space.
Hint: Lack of extreme points in BCo unlike in Bc(k)-
3.78 Show that the extreme points of Btx are exactly the vectors ±e2-.
Hint: Do it first in l\.
3.79 Show that -Bl^ci] has no extreme points.
Hint: For c with JQC |/| dt = 1/2, put /i = 2/ • X[o,c] and f2 = 2/ • X[c,i]-
3.80 Show that the extreme points of #?^[0,1] are exactly functions for
which |/|=1 almost everywhere.
Hint: Use the definition.
3.81 Show that none of C[0,1], Co, ?i[0,1] are isometric to a dual space.
Hint: Isometry preserves extreme points; the Krein-Milman theorem.

100 3. Weak Topologies
3.82 Show that C([0,1] U [2,3]) is not isometric to C[0,1].
Hint: The Banach-Stone theorem and the connectedness of [0,1], resp.
[0,1]U[2,3].
3.83 Use the Banach-Stone theorem to show that c® c is not isometric to
c. Recall that these spaces are isomorphic.
Hint: c 0 c is isometric to the space of continuous functions on the disjoint
union of two copies of one-point-compactifications of N. This set has two
limit points, c is isometric to the space of continuous functions on the
one-point-compactification of natural numbers. This set has only one limit
point.
3.84 Let K be a compact metric space such that the identity Ik is the only
homeomorphism of K onto K. Show that all linear isometries of C(K) onto
C(K) are ±IC(K).
Such K do exist. In fact, there is a connected compact K in R2 such
that the only continuous maps K —> K are constants and the identity.
Hint: The Banach-Stone theorem.
3.85 Show that ?i(c) is isometric to a subspace of C[0,1]*, where c denotes
a set of cardinality continuum.
Hint: Given (at) E ^i(c), consider the functional ^at<5(tf).
3.86 Find an example of a compact convex set C in R3 such that extreme
points of C do not form a closed set. Can that happen in R2?
Hint: Let D denote the disk of radius 1 in the (x, t/)-plane centered at
A,0,0). Let L be the line segment on the z-axis from the point @,0,-1)
to the point @, 0,1). For the set C, take conv{D U L}. Draw a picture to
see it. This cannot happen in R2, since if x is not an extreme point, then
points close to x are not extreme as well.
3.87 (Troyanski-Lin) Let x E Ext(Bx) and assume that at x the
relative norm and weak topology of Bx coincide. Show that the slices form a
neighborhood base of the norm topology at x.
Hint: Show that x is an extreme point of Bx** • To this end, assume that
x = \{x\ + X2) with Xi E Bx**. By a geometric argument, show that the
u>*-topology and the norm topology oi Bx** coincide at X{. By Goldstine's
theorem, X{ E Bx, a contradiction. Having shown that x is an extreme
point of the second dual ball, use Choquet's lemma.
3.88 Show that Bj^ does not have arbitrarily small slices.
Hint: Given F E S^ and e > 0, consider the slice S? = {x E Bt^; F{x)>
1 — e). We claim that diamEe) > 2. To this end, it is enough to prove
that S? contains at least two extreme points of Bi^. Note that functionals

3. Weak Topologies 101
not attaining their norm are dense in Bi*^ (if there was a ball of norm-
attaining functionals, all would attain and t^ would be reflexive). Hence
we may assume that F does not attain its norm.
By Exercise 3.72, S? contains at least one extreme point x. If it was the
only one, it would be in all slices S$ for 8 G @,?). Then F(x) = 1, so F
attains its norm, a contradiction.
3.89 Let X be a Banach space and C be a bounded set in X. Show that
C has arbitrary small slices if and only if conv(C) does.
Hint: Assume that C has small slices. Pick a small ?-slice C\ of C that
results from a cut of C by a hyperplane H and put D\ — conv(C \ Ci),
Ci = conv(Ci). Then convC C conv(C2 UDi). Show that if you move H
in the parallel way, the cuts of conv(C2 UDi) eventually have diameter
smaller than 2e. Details can be found in [Bou].
3.90 Find a compact convex set C in a Banach space such that C is not a
convex hull of its extreme points.
Hint: Put C — conv({2_ze;} U {0}), where e* are the standard unit vectors
in ?2. Then c = ?4-*^ ? C. By Theorem 3.41, Ext(C) C {2"iei} U {0}.
Therefore, conv(Ext(C)) is formed by points that are all finitely supported,
and thus it does not contain c.
3.91 Let Y be a closed subspace of a Banach space X. We know that Y*
is isometric to X*/YL. Show that for every x E Ext(JBx*/yx) there exists
x G x such that x G Ext(?x*).
Hint: Exercise 3.70.
3.92 Consider X = R2 with the dual norm on X* being the norm whose
unit ball is
?> = B((l,0),10)nJB((-lH),10))n{(a;,2/)€R2; -l<y<l},
where B(x, r) is a closed ball centered at x with radius r.
Prove that B = Ext(D \ {upper left corner}) is a James boundary of X
that is not closed.
3.93 Let X be a Banach space, and let B be a separable subset of X* such
that B / Bx*. Can we conclude as in Theorem 3.46 that B is not a
James boundary of XI
Hint: No. Let X be R2 with the maximum norm, and let B — {±ei, ±e2},
where e\ are the standard unit vectors in R2. Then, for each x G Sx, we
have max(#) = 1.
b v y
3.94 Show that if B is a James boundary, then Ext(Bx*) C B
Hint: By separation, conv^ (B) — Bx* • Then use Theorem 3.41.

102 3. Weak Topologies
3.95 Let V be a vector space. Recall that a map (p: V —> V is called affine if
there is a linear map T: V —> V and a vector xq such that (p(x) — xo-\-T(x)
for all x EV.
Prove that cp is an affine map of V into V if and only if tp( J2 ^ixi) —
n n
Y^, )<i<p(xi) for all x\)..., xn G V and scalars Az- such that J^ A* = 1.
2 = 1 2=1
In particular, if <p is an affine map and K C V is convex, then <?>(iv) is
convex.
Hint: If cp is affine, we have xo = ^ ^ixo and thus
n n n n n
2 = 1 2 = 1 2 = 1 2 = 1 2* = 1
If tp has the stated property, it is enough to show that ip(x) = (p(x)-(p(O)
is a linear map. For x, y G V and scalars a, /?, we have
ip(ax + Cy) = ip(ax + f3y + A - a - CH)
= <p(ax + 0y+ A - a - /?H) - <p(Q)
= ap(x) + /fy>(y) + A - a - /3)<p@) - tp@)
= aip(x) + /3<p(y) - acp@) - P<p(Q) = aip(x) + /?^(y)-
3.96 Let C be a bounded convex set in a Banach space X such that
every continuous linear functional on X attains its supremum over C. Is C
necessarily closed?
Hint: No. Consider in the plane {% ^ne convex hull of two circles of
radius one centered at B,0) and (—2,0), and remove the points (±2,±1).
The resulting set is bounded and not closed, and one can show that every
bounded linear functional attains its supremum over this set (level sets of
linear functional are lines in R2).
3.97 Let C be a weakly closed set in a Banach space X. Assume that every
continuous linear functional on X attains its supremum over C Is C weakly
compact?
Hint: Yes, see the proof of Theorem 3.58.
3.98 Find a Banach space X and a convex continuous bounded below
function / on X such that lim (f(x)) = oo and / does not attain its
||:c||-* oo
infimum on X. Can X be reflexive? Can X be finite-dimensional if we drop
the requirement on the limit of /?
Hint: Let X = Co and let / be in Sx* such that / does not attain its norm.
Define <j> for x G X by <j>{x) = ||#||2 — f(x). Then <j> is bounded below.
Indeed, if ||z|| > 2, then </>(x) > \\x\\2 - \\x\\ > 2. If ||s|| < 2, then \</>(x)\ <
Ikll2 + |/(x)| < ||z||2 + 2||/|| < 6. Let \\x\\ < \ be such that \ > f(x) > \.

3. Weak Topologies 103
Then (j)(x) — ||#||2 — f(x) < \ — f(x) < 0. Hence </> attains negative values at
some points of X. Assume that </> attains its minimum on X at a?o- Because
(f)@) = 0, we get x0 ± 0. <j)(x0) < <f>(x) reads ||z0||2 - f(x0) < \\x\\2 - f(x)
for all x E X, and hence f(x) < f(x0) + \\x\\2 — \\x0\\2 for all x E X. If
x E X is such that \\x\\ < \\xo\\, then f(x) < f(xo). Thus, / attains its
maximum on the set {x EX] \\x\\ < ||#o||}- Hence / attains its maximum
on Bxy a contradiction. The space X cannot be reflexive, because on such
a space the weakly lower semicontinuity of <j> can be used to produce the
minimum of <j> by using the limit condition on (f>. The function ex on R can
be used to finish the solution of this exercise.
3.99 Find an example of an incomplete infinite-dimensional normed space
X such that every / E X* attains its norm ([Jam3]).
Hint: Let X — ( J2 ^oo) • Note that if x = (xn) E X is an extreme
point of Bx, then xn is an extreme point of Bin for every n, so xn =
(±1, ±1,..., ±1). Let Z = span (Ext (i?x)) • X is reflexive, so by the Krein-
Milman theorem we have Bx — conv(Ext(I?x)) and thus Z — X.
Note that Z ^ X. To see this, consider the following x — (xi) E X: Xi =
m
13 X2 = (|51), ?3 = (|, |, ^), Assume that x ? Z] that is, x = J2 aiVl',
i=i
m
where y1, y2,..., ym E Ext(Bx)- Let n > 2m. Then xn = ? o^. Since
/=i
every coordinate of each vector t/^ is ±1, we have at most 2m different
coordinates of the vector xn. Since n > 2m, at least two coordinates of xn
should be equal, which is not the case. Hence x ? Z.
Now pick any / E Z*. Extend it on X and call it / again. Because Bx
is weakly compact, / attains its norm on Bx- By the proof of the Krein-
Milman theorem, we have that / attains its norm at an extreme point of
Bx- Using the fact that the norm of / on X and Z are the same, we get
that / attains its norm on Bz-
3.100 Show using Theorem 3.55 that BCo is not weakly compact.
Hint: B~*) E t\ does not attain its norm on BCQ. Indeed, if x E BCo then
eventually |a?t-| < 1, so |E2~f^l < E2"' = L
Direct proof: i E eM nas no ^-cluster point.
z=1
Note that exactly finitely supported elements in t\ attain their norms.
3.101 Show that the functionals in c? that attain their norms do not form
a residual set in Cg.
Hint: Recall that it is exactly finitely supported functionals that attain their
suprema over BCo (see the previous exercise). Then, find the Baire category
of the set of finitely supported vectors. Use the fact that finite-dimensional
subspaces of t\ are closed and have empty interiors in l\.

104 3. Weak Topologies
3.102 Let X be a Banach space and / G Sx*- Does there exist fn G X*
that attain their norms such that ||/n — /|| —» 0 and fn lie on one line
through /?
Hint: Not in general. Consider in cj = l\ an element with infinite support,
and use the fact that precisely finitely supported vectors in l\ attain their
norms.
3.103 Assume that C is a separable closed convex subset of X* such that
each x G X attains its supremum over C. Is C u?*-compact?
Hint: Yes, by the proof of Theorem 3.46, C is equal to its w*-closure.
3.104 Show that a Banach space X is reflexive if and only if the distance
of any point to a given closed convex subset of X is attained.
Hint: Assume there always is a closest point. Let d = dist(#, C) and cn G C
be such that \\x — cn\\ —> d. Then, since cn form a bounded sequence and
the space is reflexive, there is a subsequence cnk such that cnk —> c in X
(Exercise 3.111). Since C is ^-closed, we have c G C. The norm is weakly
lower semicontinuous because the sets || • ||~ (—oo, a] are convex and closed
and thus weakly closed. Hence \\x — c\\ = d.
If X is not reflexive, then by Corollary 3.56 there is an element / G Sx*
such that / does not attain its norm. Then yo@,/_1(l)) = 1 < ||x|| for
every x G /_1A).
3.105 Prove the following Smulian's theorem: A Banach space X is
reflexive if each nested sequence Cn D Cn+i of closed convex subsets of Bx has
a nonempty intersection.
Hint: Show that then every / G X* attains its norm by considering the
sets Cn = {x G Bx ; f(x) > sup(/) - ?}.
BX
3.106 Let X be a Banach space, and let C be a separable set in X*. If C
is w*-compact, is D = conv(C) also w*-compact?
What if the separability assumption is dropped?
Hint: Yes for the first part. Each x ? X attains its supremum over C which
is the supremum of x over D. Then use Exercise 3.103.
If the separability is dropped, the situation is different. Consider the
Dirac measures in ?c[o,i]*; see Exercise 3.73.
3.107 Let X be a separable Banach space, and let B be a James boundary
for X. Show that if A C X is a bounded set in X compact in the topology
of pointwise convergence on B} then A is tu-compact.
Hint: We may assume that B is convex. Then B = Bx*- Because Bx*
is metrizable in the w*-topology, B is ^-separable and there is a
countable w*-dense subset D of 5, in particular D = Bx* • The topology of

3. Weak Topologies 105
the pointwise convergence on D is a metrizable topology on A, which, by
the compactness assumption, coincides with the pointwise topology on B.
Thus, every sequence in A has a subsequence that is convergent in the
topology of the pointwise convergence on 5, which by Theorem 3.60 gives
that A is weakly compact.
3.108 (i) Show that Theorem 3.60 does not in general hold for nets.
(ii) Show that Theorem 3.60 does not in general hold for unbounded
sequences.
(iii) Show that Theorem 3.60 does not in general hold for x E X**.
Hint: (i): For finite F C [0,1], consider the following function fp E Sc[o ,!.]•'
assume F — {#;}, 0 < x\ < ... < xn < 1. Define fp as the piecewise
linear function with nodes in xi) where /f(^») = 1, and nodes in 0,1, and
midpoints of [x{, Xi+i], where fp = 0. Then for the Lebesgue measure A we
have A(/p) = |, but considering the net {/f}, F partially ordered by the
inclusion, we see that <$*(/f) —> 1 for all Dirac measures.
(ii): Put xl — iei, where ez- denote the standard unit vector in Co- Then
xl —> 0 pointwise in Co, but xl does not converge weakly to 0 in c$ because
it is unbounded.
(iii): Let F E C[0,1]** be zero on all Dirac measures but F ^ 0. Take
a?n = 0GC[0,l].
3.109 Show that Lemma 3.47 does not hold in general if the assumption
on the completeness of X is dropped.
Hint: Consider the subspace X of t^ formed by all sequences that are
eventually constant. Then the standard unit vectors {e2-} form a James
boundary B of X. Consider xl = X{*,*+i,...}- Then supB(limsup(a?z)) = 0.
If x E conv<(V}, then for large n, the n-th coordinate of x equals 1. Note
that this is caused by the fact that infinite convex combinations are used
in the proof of the Simons inequality.
3.110 Show that Bi* is not w*-sequentially compact.
Hint: Define a sequence of elements of Bi* by fn ((#«)) = xn. This sequence
has no w*-convergent subsequence. To see it, assume that {fnk} is a w*-
convergent subsequence. Define a vector x = {x{) E ^oo by xnk = (—l)k
and X{ — 0 if i ? {n^}. Then fnk(x) is not convergent.
3.111 (i) Let {fn} be a sequence in the dual X* of a Banach space X.
Show directly that if X is separable and {/n} is bounded, then there exists
a w*-convergent subsequence of {/n}.
(ii) Let {xn} be a sequence in a Banach space X. Show directly that if X is
reflexive and {xn} is bounded, then there exists a tu-convergent subsequence
of {xn}.
Hint: (i): Let {x{} be a dense countable subset of X. Since {fn} is bounded,
so is {fn(xi)} and there exists a convergent subsequence {/^(#i)}. Apply-

106 3. Weak Topologies
ing this trick to {/^} and x2, we get a subsequence {/^} such that {fn(xi)}
converges for i = 1, 2. Continue in this manner and set fnk = /?. Then, for
every Xi, the limit lim(/nfe(?;)) exists.
Using the boundedness of {fn} and density of {#z}, show that
lim(fnk(x)) exists for all x G X\ call it f(x). f is linear, and assuming
||/n|| < C? we show that |/(#;)| < C||?i||, so / is continuous.
(ii): Define Y = span{^n}. Then Y is reflexive and separable, hence also
y** = y and consequently Y* are separable. Since {xn} is bounded in
y**, by the first part it has a subsequence {xnk} that W*-converges in y**
to some x G Y** = Y. Thus f(xnk) —» f(x) for all f EY*y and hence also
for all / G X* and ?nfc ^ x.

4
Locally Convex Spaces
In this chapter, we restrict ourselves to the real scalar field.
Definition 4.1
Let E be a vector space, and let r be a Hausdorff topology on E such that the
operations (x,y) E E xE h-> x + y and (x,a)G^xR^ax are continuous
on Ex E and ExTl, respectively. Then (?", r) is called a topological vector
space.
Therefore, for every #i, x^ € E and a neighborhood W of x\ + #2, there
are neighborhoods V\ and V2 oix\ and ?2, respectively, such that V1 + V2 C
W, where Vi 4- V2 = {# -f y; x E Vi, y E V2}. Also, for every x E E, a ?H
and a neighborhood W of a#, there is a neighborhood V of a? in E and
6 > 0 such that j3V CW for every |/3 - a| < 6, where /?V = {/?v; v E V"}.
We have the following statement.
Fact 4.2
Zetf X 6e a topological vector space.
{%) For every a E E, the translation operator Ta defined for x E E by
Ta(x) — x + a is a homeomorphism of E onto E.
(ii) For every a E R, ol ^ 0, tfAe multiplication operator Ma defined for
x E E by Ma(x) = a# is a homeomorphism of E onto E.
By a local base for a topological vector space (E, r) we mean a local base
at the origin 0; that is, a collection B of subsets of E such that each U E B
contains a neighborhood of 0 and for every V E r with 0 E V we can find
17 E B such that 0 E U C V\

108 4. Locally Convex Spaces
Lemma 4.3
// B is a local base of a topological vector space E, then every set from B
contains the closure of some set from B.
In particular, E has a local base consisting of closed sets.
Proof: Let U G B. Since (#, y) »—> x — y is continuous, we find a
neighborhood V of zero in E such that V - V C U\ that is, V n ((E \ U) + V) = 0.
Since (E \ U) + V is an open set, we have V C\((E\U) + V) = 0; in
particular, Ffl (? \ J7) = 0 because 0 G V". Thus VcU.
D
Definition 4.4
A topological vector space E is called a locally convex space ifE has a local
base consisting of convex sets.
A topology r on E is called a locally convex topology if (E, r) is a locally
convex space.
The norm, w-, and w*-topologies are examples of a locally convex
topology.
A metric d(-, •) on a vector space E is called (translation) invariant if
d(x + z,y+ z) = d(x, y) for all x,y,z G E.
Definition 4.5
A locally convex space E is called a Frechet space if its topology is induced
by a complete invariant metric.
A locally convex space E is called normable if its topology is induced by a
norm on E.
In particular, if X is a Banach space, then the metric d(z, y) — \\x — y\\
makes it a Frechet space.
Definition 4.6
A set A in a vector space E is called balanced if a A C A for every \a\ < 1.
Proposition 4.7
(i) Every topological vector space E has a local base consisting of balanced
sets.
(ii) Every locally convex space E has a local base consisting of convex and
balanced sets.
Proof: (i): Let U be a neighborhood of 0 in E. By the continuity of scalar
multiplication, there is 6 > 0 and a neighborhood V of 0 in E such that
aV C U for every |a| < S. Put W = (J aV. Then W is a balanced
neighborhood of 0 in E and W C U.
(ii): Let U be a convex neighborhood of 0 in E. First, we construct
W as in (i). Since W is balanced, then V = conv(W) is a balanced (see

4. Locally Convex Spaces 109
Exercise 4.1) and convex neighborhood of 0, and V C U because U is
convex.
?
Proposition 4.8
Let E be a topological vector space. If dim(E) = n, then E is linearly
homeomorphic to P^.
Proof: Let {ei,..., en} be a basis of E. We define a map u from ?% on^°
n
E for x — (x{) by u(x) = J2 xiei-
By the continuity of vector operations in E, u is a continuous linear
bijection of P% onto E. To complete the proof, we must show that w_1
is continuous. Let B\ denote the unit ball of ?%] we will show that u{B\)
contains a neighborhood of 0 in E. By the linearity of u and of the topologies
of E and P\, this will prove continuity at all points of E.
Let S\ be the unit sphere in P%. Because S\ is compact in P^ and u is
continuous, u(Si) is compact in E. Also, u is one-to-one, so we have that
0 ? u(Si). Since E is a HausdorfF space, for every s E u(Si) there are
neighborhoods Us and Vs of s and 0, respectively, such that Us C\VS / 0.
By compactness, there are si,..., sp E u(S\) such that u(Si) C (J ?/*,.
2 = 1
P
Put V = p| V^-. Then V is a neighborhood of zero in E such that
2 = 1
V fl w(S'i) = 0. By Proposition 4.7, it follows that there is a balanced
neighborhood W of zero in E such that VP C V and thus W fl u(Si) = 0.
We claim that W C u(Bi). Indeed, if for some w EW we have w ? u(J3i),
then w = u(v) for some v ^ B\. Then |Ar E Si and u(tt^) = tAt E VF,
because W is balanced, w E TV and ||i;|| > 1.
?
Recall that a net {xa}aei in a topological vector space E is called Cauchy
if for every neighborhood [/ of 0 in E there is ao E J such that {xa—xp) E Z7
whenever c*o < a, f3. E is called complete if every Cauchy net is convergent.
Corollary 4.9
Let F be a subspace of a topological vector space E. If F is finite-
dimensional, then F is closed in E.
Proof: Since ?% is a complete normed space, by Exercise 4.7 it is also
complete as a topological vector space. By Proposition 4.8, the same is true
for F. Let y E F. Then there is a net {xa} C F such that lim(ara) = V-
Clearly, {xa} is a Cauchy net in F, so it converges to some element in F.
Hence y E F.
D

110 4. Locally Convex Spaces
Proposition 4.10
Let E be a locally convex space. If E has a countable local base, then the
topology of E is induced by a translation-invariant metric.
Proof: Let {Un} be a countable local base in E. By Proposition 4.7, we
can assume that every Un is a convex and balanced set. For every n, let
pn be the Minkowski functional of Un. Then pn is a seminorm; that is,
Pn(x + y) < Pn(x)+Pn(y) for x,y e E and pn{ocx) = Hp(e) for x ? E and
a E R.
Define a metric d on E by
«->=?rj&
To verify that d is indeed a metric, note that j^ is increasing. Thus, for
x,y,z E E,
vY
l-ft
pn(x-z) pn(x - y) + pn(y - z)
l+Pn(x-z) ~ l+pn(x- y)+Pn(y- z)
_ Pn(x-y)
l + Pn(x-y)+Pn(y- z)
Pn(y-z)
+
l + pn(x-y) + Pn(y- z)
< Pn(x - y) Pn(y~z)
l + Pn(x-y) l+Pn(y-z)'
Since the series in the definition of d is uniformly convergent for x, y G E,
we have that the balls {x E E\ d(#, 0) < r} are open in E. The metric d
is translation-in variant, so to prove that d induces the topology of E, it is
enough to show that every Un contains the ball {x ? E\ d(x, 0) < 2~(n+1)}.
If d(^,0) < 2"(n+1), then „ Pn, x < i and since the function ^ is
l+Pn(x) l 1+t
increasing, we have pn(x) < 1 and thus x ? Un.
n
Definition 4.11
Let A be a subset of a topological vector space E. A is called bounded if for
every neighborhood U of 0 in E there is a > 0 such that ocA C U.
A is called totally bounded if for every neighborhood U of zero in E there
is a finite set F C E such that A C F -f U.
Note that every totally bounded set A in a topological vector space E is
bounded. Indeed, given a balanced neighborhood W of zero in E, we use
the continuity of the addition to obtain a balanced neighborhood V of zero
in E such that V-f V C W. Since A is totally bounded, there is a finite set
F C E such that A C F + V. Let F — {xly..., xn}. From the continuity of
the scalar multiplication, we find 1 > 6{ > 0 such that 6x{ ? V whenever

4. Locally Convex Spaces 111
0 < 8 < Si. Set 8 = min{<!>;}; then 8xi E V for every i. From this and the
balancedness of V, we obtain 8 A C 8F + 8V C V + V C W. This shows
that A is bounded in E.
Proposition 4.12
Let E be a locally convex space. E is normable if and only if E has a
bounded neighborhood of zero.
Proof: Clearly, if E is normable, then it has a bounded neighborhood of
zero, namely the open unit ball centered at the origin. Assume that U is a
bounded balanced convex neighborhood of zero in E. Then the Minkowski
functional p of U is a continuous norm on E which induces the topology of
E. Indeed, if V is a neighborhood of zero in Ei then all C V for some a > 0,
showing that the balls centered at the origin and generated by p form a
local base of E. To check the continuity of P, we use \p(x)—p(y)\ < p(x — y))
which follows from the triangle inequality.
?
Proposition 4.13
Let E be a topological vector space. If E has a totally bounded neighborhood
of zero, then E is finite-dimensional
Proof: Let U be a totally bounded neighborhood of 0. It is also bounded,
hence {2~nU} « forms a local base of E. By the total boundedness of
U, there is a finite set A C E such that U C A + \XJ. Let F = span(yl).
Then U C F+ \U. Multiplying this inclusion by |, we have \U CF+\U.
Combining these inclusions, we have U C F + F + ^U — F -f |{7. By
induction, we obtain U C F + 2~nU for every n. Therefore, U C f](F +
2-nU)C~F.
The last inclusion follows from the fact that {2~nU} forms a local base
in E. Since F is closed in E by Corollary 4.9, we have U C F. Given x ? E,
we have 0 • x E U; thus there is 8 > 0 such that 8x C U C F and hence
x E F. Therefore F = E, and E is finite-dimensional.
?
Definition 4.14
Let E be a topological vector space. The space E* denotes the set of all
continuous linear fundi on als on E.
Lemma 4.15
Let E be a topological vector space, and let f be a linear functional on E.
The following are equivalent:
(i) f is continuous on E; that is, f € E*.
(ii) f is continuous at some point of E.
(Hi) There is a neighborhood U of some xo E E such that f is bounded
above on U.

112 4. Locally Convex Spaces
Proof: The implications (i) => (ii) =>• (iii) are trivial.
(iii) => (i): Let C > 0 be such that f(x) < C for all x E U. Set
D — C — f(xo) and V = ^(U — xo)] we may assume that V is balanced.
Then V is a balanced neighborhood of 0 and f{x) < 1 for all x ?V. Given
x E V, we have —z E V, so /(#) > — 1 as well, that is, |/| < 1 on V.
Now given y ? E and ? > 0, W = y + eV is an open neighborhood of y
such that |/(#) — f(y)\ < ? for x E W; that is, / is continuous on i?.
?
The following example shows that unless the topology of E is locally
convex, the space E* may be degenerate.
Example
Fix p E @,1). Let Lp denote the vector space of Lebesgue-measurable
functions on [0,1] for which
9(f) = I |/@IP *< oo.
JO
Since for a > 0 and b > 0 we have (a + b)p < ap + bp, it follows that
q(f + 9) < q(f) + ?(#)• Therefore, the formula d(/,flO = q(f - 9) defines
a translation-invariant metric on Lp in which Lp is complete. This follows
in the same way as for Lp, p > 1. We claim that if O is a nonempty open
convex set in Lp, then O = Lp.
To prove this claim, assume that V ^ 0 is an open convex set in Lp such
that 0 E V. Choose / E Lp. We will show that / E 7. Choose r > 0 such
that Br C V, where Br = {f ? Lp] q(f) < r}; then choose n E N such that
np~lq(f) < r. By the continuity of the indefinite integral J* \f\p dt, there
are points 0 = xq < x\ < • • • < xn = 1 such that J * \f\p dt — n~1q(f)
for i — 1,... , n.
Define for z = 1,..., n functions gi on [0,1] by
d\ _ / nf(t) f°r x%-i < i < 3»"
^W " \ 0 otherwise.
Then # E V since q(gi) = //.^ |n/|* A = np~lq(f) < r and 5r C 7.
Since V is convex and / = ^{91 -f h #n), it follows that / E V and
the claim is proved.
Consequently, the only open convex neighborhood of 0 is the whole Lpj
so the topology is not locally convex.
Also, let F E L* Because ir'~1((—a,a)) is a convex neighborhood of
zero in Lp for every a > 0, we have F_1((—a,a)) = Lp. Thus F = 0, and
1; = {o}.
Example
Let Q, be an open set in Rn. We will now construct the space of distributions
and establish some of its properties.

4. Locally Convex Spaces 113
Choose any sequence {ifn}n€N of compact subsets of Q such that
Kn C Int(if„+i) and |J Kn = fi; for example, Kn = {x] dist(z, Rn \ Q) >
^Nl2<n}.
Definition 4.16
Let C(Ct) denote the vector space of all real-valued continuous functions on
Q, with the topology r determined by the local base {Un}, where
Un = {f€ C(fi); sup |/(*)| < i}.
*?Kn
Fact 4.17
(i) (C(Q),r) is a locally convex space whose topology does not depend on
the choice of{Kn}.
(ii) The topology r is the topology of uniform convergence on every compact
subset ofQ; that is, fa—*f in r if and only if fa —> / uniformly on every
compact subset ofQ.
(Hi) It is a Frechet space but the topology r is not normable.
Proof: Local convexity of r is routine to verify.
Let {Kfn} be another family of compacts with the required properties.
Then, for n G N, {lnt(Km)}m is an open cover of K'n) and hence, using
Kn C Int(ifn+i), there is Km such that Kfn C Int(Km) C Km.
This shows the independence of r on the choice of {Kn} and also that r
is the topology of uniform convergence on every compact in ?1.
By a standard argument, we see that (C(fi), r) is a Frechet space. Since
every Un contains functions / for which sup |/| can be arbitrarily large,
we have that no Un is bounded. Therefore, (C(Q), r) is not normable.
?
In particular, any set of the form < f G C(fi); sup \f(x)\ < s\ with e > 0
^ xeK J
and K a compact in fi is an open neighborhood of 0 in C(Q).
Recall that for every multiindex a = (ai,...,an), where <y.{ are
nonnegative integers, the corresponding derivative is defined by
The degree of the derivative is defined as \a\ — Y2ai-
Definition 4.18
By C°°(fi) we denote the set of all real-valued functions f on Q such that
Daf G C(Q) for every a. A topology on C°°(Q) is defined by its local base
{Unta}
^n,a-{/EC°°(fi); sup \Daf(x)\ < -).

114 4. Locally Convex Spaces
Note that the base {UnyQ} is countable.
Fact 4.19
(i) C°°(Q) is a Frechet space whose topology does not depend on the choice
of{Kn}.
(ii) fp —> f in C°°(fi) if and only if Da fp —> Da f uniformly on K for
every compact K in Q and every multiindex a.
(Hi) C°°(fi) has the Heine-Borel property.
Recall that a space has the Heine-Borel property if all of its closed
bounded subsets are compact.
Proof: We will show (iii). Let Abe a bounded and closed set in C°°(Q).
Then, for every rc, there is an Mn > 0 such that \Daf(x)\ < Mn for all
\a\ < n? / ? A an<i x E Kn. Therefore {Da/; / E A} is equicontinuous on
Kn-i for \&\ < n—1. By the Arzela-Ascoli theorem and the Cantor diagonal
process, it follows that every sequence {/»•} C A contains a subsequence fa.
for which all DQ/,-. converge uniformly on every compact subset of Q.
Hence, /«. is convergent in C°°(Q). This proves that A is compact.
?
The differentiation mapping Da:C°°(fi) —> C°°(Q) is continuous and
linear.
Recall that if / is a real-valued function on fl, the support of / is defined
by supp(/) - {x E fi; f(x) ± 0}.
Definition 4.20
Let K be a compact set in fl such that lnt(K) ^ 0. We define T>k as the
closed subspace o/C°°(Q) formed by functions f such that supp(f) C K.
We define V{Q) as the union of allVxiJl) for compacts K as above.
Note that V(Q) is not a complete subspace of C°°(Q). For Q, = R, this
is seen as follows. Choose <p E C°°(R) with supp(<?>) C [0,1] and <p > 0 on
@,1). For mEN, put
<Pm(x) = <p(x - 1) + \<p{x - 2) + h ±(p(x - m).
Then cpm is a Cauchy sequence in V(R) C C°°(fi) and lim(<?>) does not
have a compact support.
We now introduce a locally convex topology on X>(Q).
For every compact K C fi with nonempty interior, let tk denote the
topology of T>k inherited from C°°(Q). Let B denote the collection of all
convex balanced sets W C V(Q) such that T>k fl W is r^-open in T>k for
every such compact K C fi. Then 6 is a local base for a locally convex
topology on T>(Q), which will be denoted by r.
Proposition 4.21
(i) The topology on T>k inherited from V(Q) coincides with tk for every
compact set K C ?1 with Int(K) / 0.

4. Locally Convex Spaces 115
(ii) If A is a bounded subset ofV(Q), then there is a compact K CQ, such
that ACVK.
[Hi) V(Q) has the Heine-Borel property.
(iv) Every Cauchy sequence in V(Q.) is convergent.
(v) V(Q) is not metrizable.
Proof: (Sketch) To prove (i), it is enough to show that, given an open set
O in T>k, there exists an open set V in V(Q) such that O = V 0 T>k- Let
{Un} be the local base in T>k as above and pn be the Minkowski functional
of Un. For every <p € O, there is n and 8 > 0 such that
{il>evKi Pn(*P-<p)<6}cO.
Put Wy = {i/>e 2>(fi); PnW < S}. Then W^ G B and
vKn(<p + w(p) = <p + (vKn wv) c o.
Let V = |J (cp + W<p). Then V has the desired property.
vet)
(ii): Assume that A is bounded in V(Q) and A lies in no ?>#• Then there
are (pm G A and distinct points #m G fi without a limit point in Q, such
that (pm(xm) / 0, m G N. Let Wbe the set of all (p G V(Q) that satisfy
|^(«m)| < ^|y>m(sm)| for all m G N.
Note that W^ is convex and balanced. Since every compact K contains
only finitely many xm, it is easy to see that Vk^W is open in T>k • Therefore
W G B. Since <pm ? mW, no multiple of A is contained in W. Therefore,
A is not bounded in X>(fi).
(iii) follows from (ii) and the fact that T>k has the Heine-Borel property.
(iv) follows from the fact that every Cauchy sequence is bounded, from
(ii) and from the completeness of T>k •
(v): It is easy to see that T>k has no interior point in V(Q). Assume that
V{Q) is metrizable. Then V(Q) is a complete metrizable space because any
Cauchy sequence in V(Q,) is convergent. Note that Vj. are closed subspaces
of X>(Q). This leads to a contradiction with the Baire category theorem.
?
We now characterize continuity of linear maps on V{Q).
Proposition 4.22
A linear functional f on ?>(fi) is continuous if and only if f is sequentially
continuous if and only if the restriction of f to every T>k is continuous.
Similarly, a linear map T:X>(fi) —» E, where E is a locally convex space, is
continuous if and only s/TL is continuous for every Vr.
PROOF: Assume that the restrictions of / to all T>k are continuous.
Consider the set V = /_1((—6,6)). Then V is balanced and convex. By
Proposition 4.21, V is open in V(Q) if and only if T>k H V is open in
T>k for every K. But this is the case since the restriction of / to each T>k
is continuous.

116 4. Locally Convex Spaces
If / is sequentially continuous on ?>(Q), then the restriction of / to every
T>k is continuous because T>k is metrizable for every K.
?
Definition 4.23
Continuous linear funciionals on V(Q) are called distributions. It is
traditional to denote V(Q,)* by V'(?l).
We will now show some important examples of distributions.
The space of locally integrable functions can be embedded into V'{Q).
Indeed, let / G Lloc(Sl). We define Tf(<p) = fKnf<pduj. Then 7) is a
distribution. The continuity on every T>k follows from
/ ftp\du) < / |/|max|y>| du.
Jk I Jk k
In particular, we may consider C(Q) a subset of V'(Q).
The space of all Borel measures on Q can be also embedded into V'(Q).
It is easy to see that the functional T^np) = JRn cpdp, is a distribution. In
particular, Dirac measures are distributions.
Let a be a multiindex. For F G V'(Q), we define the derivative by
DaF:<p~(-l)MF(Da(<p)).
It is easy to see that DaF G Z>;(fi).
The definition is motivated by the following. Let T be the distribution
corresponding to some function /; that is, T = Tj. Assume that / has
continuous derivatives of order |a|. Using integration by parts and the fact that
<p G V{Q) vanish on the boundary of fl, we obtain DaTf((p) = Td°j(<p) for
all (p G V(Q). Thus for differentiate functions, the distributional derivative
Da agrees with the usual partial derivative.
Fact 4.24
Da:V\^i) —>?>'(?}) is a continuous linear operator.
Proof: Linearity is obvious.
Let K be a compact set in ?2, and let F G V\Cl) be bounded by C > 0
on a neighborhood Un from the local base in T>k- Let pn be the Minkowski
functional of Un. Then |i^(^)| < Cpn((p) for every ip G T>k- Consequently,
\(DaF)(<p)\ < Cpn{D°v) < CPnMa\W)-
Therefore, DaF is a continuous linear functional when restricted to an
arbitrary T*k> which by Proposition 4.22 means that DaF G V'(Q).
n
The space Vl(Q) is important in the theory of generalized solutions to
differential equations. Note that every continuous function on Cl has all
partial derivatives in the sense of distributions.

4. Locally Convex Spaces 117
We will now return to the theory of locally convex spaces and start with
a separation result.
Theorem 4.25
Let C be a closed convex set in a locally convex space E. If xq ? E\ C,
then there is f ? E* such that f(xo) > sup{/(#); x ? C}.
Proof: By considering a shift of C, we may assume that 0 ? C. Let U be
a convex symmetric neighborhood of 0 such that (x0 + 2U) fl C = 0. Then
(so + U) H (C + U) = 0, and thus setting D - C + U, we have x0 g ~D- Let
fi be the Minkowski functional of D. If /jl(xo) < 1, then xo ? (l + ~)-D for
every n ? N and thus A +i) z0€ ?>,(! + ?) #o —^^o- Hence xq ? D,
a contradiction. Thus fi(xo) > 1.
Define a linear functional / on span{#o} by f(axo) = a/i(xo). We claim
that / < /i. Indeed, for a > 0 we have f(ax0) = a//(x0) = //(aa?o)> and for
a < 0 we get f(axo) = a/z(a?o) < 03 whereas /i(aa?0) > 0 by the definition of
/x. By the Hahn-Banach theorem, there is a linear functional on E (denoted
by / again) that agrees with / on span{z0} and is dominated by \i on E.
Since / < \x on E and fi < 1 on D by definition, for every x ?U C D we
have /(#) < 1. From the symmetry of U, we get |/| < 1 on U, and hence /
is continuous on E. From C C D we also have sup{/(#); x ? C} < 1, and
/(a?0) = A^(^o) > 1.
D
The following corollary is immediate.
Corollary 4.26
Let E be a locally convex space and x,y ? E. Ifx^y, then there is f ? E*
such that f(x) ^ f(y)-
In particular, E* is nontrivial if E is nontrivial.
Definition 4.27
Let E be a vector space. By E# we denote the vector space of all linear
functionals on E.
Let F be a subset of E#. The a{E) .F)-topology on E is defined as the
topology generated by the basis consisting of sets
{x ? E) \fi(x - xo)\ < e for all i = 1,..., n}
for all choices of xq ? E, f\ ..., fn ? F, and e > 0.
Recall that a subset F C E# is said to separate points of E if for x ^
y ? E there is / ? F such that f(x) / /(y). We have just shown that, for
a locally convex space ?", the set E* separates points.
Note that if F separates points of E, then the (t(E, F)-topology is Haus-
dorff. (i?, a(E, F)) is then a locally convex space. Observe that xa —+ x in
a(E, F) if and only if f(xQ) -+ f(x) for all / ? F.

118 4. Locally Convex Spaces
Proposition 4.28
Let E be a vector space. For every F C E#, (E, <r(E, F)) — span(F).
Proof: By the definition of the a(E, F)-topology, F C (E, a(E, F))*. We
must show that given / ? (j5, cr(E, F)) there are /i,..., /n such that
f =J2 aifi- We will use Lemma 3.9.
Since / is continuous in the cr(E} F)-topology, there is an open
neighborhood V of 0 such that \f(x)\ < 1 for all x ? V. We may assume that there
are/i,...,/nGF and e > 0 such that V = {x ? E\ \fi(x)\ < e}.
Take any z such that fi(z) = 0 for all i. Then also |/2(n2;)| < e for all
n ? N. By the linearity, this means that \f(z)\ < - for all n ? N, that is,
/(z) = 0. This shows that f] Ker(/2) C Ker(/), and the proposition follows.
i
n
Note that if X is a normed linear space, then its w-topology is equal to the
cr(X,X*)-topology, and the w*-topology on X* is equal to the a(X*,X)-
topology, where X is considered a subset of (X*)#.
Similarly, if E is a locally convex space, we call the o-(E, E^-topology
the weak topology of E, and the o~(E*, i?)-topology is called the wea& 5^ar
topology of ?"\
Note that the proposition above implies that (E1*,^*)* = E\ in
particular, (X*,w;*)* = X for a Banach space X.
Repeating the argument of Theorem 3.19, we obtain the following
theorem.
Theorem 4.29
Every closed convex set in a locally convex space is weakly closed.
Definition 4.30
Let E be a vector space, and let F be a subspace of E^ that separates points
ofE. Let ACE, B C F.
By the polar of A in F we mean A0 = {/ ? F; f{x) < 1 for all x ? A}.
By the polar of B in E we mean Bo — {x ? E\ f{x) < 1 for all f ? B}.
It easily follows that A0 is a convex and <r(.F, Enclosed subset of F —
(E, <t(E, F))* for every Ac E. Similarly, B0 is a convex and a(E, F)-closed
subset of E. If A is balanced, so is the corresponding polar set. We then
have j4° = {/; sup \f(x)\ < 1}, which is sometimes used as the definition
xeA
of a polar set.
Note that {Mo; M a finite subset of E*} is a local base for the weak
topology of E\ a similar fact about the weak star topology is also true.
Theorem 4.31 (Alaoglu, Bourbaki)
Let E be a locally convex space. If A C E contains a neighborhood of 0,
then A0 is w*-compact.

4. Locally Convex Spaces 119
Proof: First, assume that A is also balanced and convex. The proof is
analogous to that of Theorem 3.21. We observe that the set [—1,1]A of all
functions / from A to [—1,1] is compact in its pointwise topology. We define
a map F: A0 —» [—1,1]A by F(f) = /| . Since A contains a neighborhood of
0, F is a bijection, and it is easy to check that F is in fact a homeomorphism
of (A0, w*) onto F(A°) in the pointwise topology of [-1,1]A.
Therefore, for the compactness it is enough to prove that F(A°) is
pointwise closed in [— 1, l]A, which is done just as in Theorem 3.21.
For a general A containing a neighborhood of 0, we find a balanced
convex neighborhood B of 0 such that B C A (Proposition 4.7). Then A0
is a tu*-closed subset of a w*-compact set f?°, and the theorem is proved.
a
In particular, consider a vector space E and a subspace F of E# that
separates points of E. If A C E contains a o~{E) .F)-neighborhood of 0, then
A0 is a(F, .^-compact in F.
Theorem 4.32 (Bipolar theorem)
Let E be a vector space, and let F be a subspace of E# that separates points
of E. Denote by a the o~(E, F)-topology on E. For every subset A of E, we
have (A°H - conv^A U {0}) {polar in F).
In particular, lei E be a locally convex space. If A C E is weakly closed,
convex, and 0 E A, then A = (^4°)o-
Similarly, if B C E* is w*-closed, convex, and 0 E B, then B = (Bo)°.
Proof: Denote C = conva(,4 U {0}). Since A C (A°Hi 0 E (A°H, and
(A°)o is cr-closed and convex, we have C C (^°)o- Assume that there is
x E (A°)o \ C.
By Theorem 4.25, there is / E F = (E,<r)* such that f(x) > supc(/).
Since 0 E C, we have supc(/) > 0, so by scaling we may assume that
supc(/) < 1 and f{x) > 1. Then supA(/) < 1, hence / E A0, and also
x E (^4°)o, so we have f{x) < 1, a contradiction.
?
We now show some applications of the bipolar theorem.
Let E be a vector space, and let F be a subspace of E& that separates
points of E. A family Q of cr(F, E')-bounded sets in F is called saturated
if Q is closed with respect to taking scalar multiples and balanced convex
a(F, i?)-closed hulls of its members, and for all Si, 52 E Q there is S3 E Q
with Si U S2 C S3.
Consider a saturated family Q and assume that |J G separates points
of E. The topology Tg on E is generated by the local base {U(G,e)\ G E
G,e > 0}, where U(G,e) = {y E E\ \g(y)\ < e for all g E G}. It is called
the topology of uniform convergence on Q. It is routine to check that Tg is
a locally convex and Hausdorff topology on E. Note that U{G) e) = eGo if
G is balanced.

120 4. Locally Convex Spaces
Theorem 4.33 (Mackey, Arens, Katetov)
Let E be a vector space and let F be a subspace of E^ that separates points
of E. A locally convex topology r on E satisfies (E, r)* — F if and only
if r is the topology of uniform convergence on some saturated family Q of
a(F,E)-compact convex balanced sets in F that covers F.
Proof: Let Q be a saturated family satisfying the assumptions. If / ? F,
then / ? G for some G ? Q. Then / is bounded on U{G) 1); hence, by
Lemma 4.15, / ? (E, Tg)*. Conversely, if / ? (E, Tg)*, then there is G ? Q
and e > 1 such that U(G,e) C /_1(—1,1). Since Q is saturated, we may
assume that e — 1 and G is balanced, so U = E/(G, 1) = Go and /L < 1.
Thus / ? U° = (GoH. Since G is <r(E, inclosed, so is G U {0}, and thus
by the bipolar theorem, / ? conv(G U {0}) C F.
Now assume that a locally convex topology r on E is such that (?", r)* =
F. Define Q — {U°; U is a balanced r-neighborhood of 0} (polar in F).
Note that U° is a o"(ir', #)-compact set in F by Alaoglu's theorem; it is also
balanced and convex. Using in particular A0 U B° C (A fl JB)°, we get that
Q is a saturated family. Clearly, F = (J G.
Gee
We will show that r = 7^. Let V be a r-neighborhood of 0. By Lemma 4.3
and Proposition 4.7, we may assume that V is r-closed, convex, and
balanced. Thus V° ? Q and V = (V°)q is open in the ?-top. Consequently,
the ?-top is stronger than the r-top.
Let V be an open ?-top neighborhood of 0. We may assume that V =
eU{G) 1), where G = U° for some balanced r-neighborhood U of 0. Then
U(G, 1) = Go = (U°)o D U and V = d7 is r-open.
?
Definition 4.34
Let E be a locally convex space. The locally convex topology on E of uniform
convergence on all w*-compact convex balanced sets in E* is called the
Mackey topology on E and denoted fi(E,E*).
By Theorem 4.33 and the note after Definition 4.27, we obtain the
following corollary.
Corollary 4.35
Let E be a locally convex space. Then the strongest locally convex topology
r on E for which (E,r)* = E* is the Mackey topology fi(E,E*) and the
weakest locally convex topology a on E for which (?, cr)* — E* is the weak
topology on E.
Similarly, the strongest locally convex topology r on E* for which (E*, r)* =
E is the Mackey topology fi(E*, E), and the weakest locally convex topology
a on E for which (i?, cr)* = E* is the w* -topology on E.

4. Locally Convex Spaces 121
Proposition 4.36
Let X be a Banach space. Then the original norm topology on X coincides
with the Mackey topology on X.
Proof: By Theorem 4.33, n(X,X*) is stronger than the norm topology.
Conversely, let 7bea fJ>(X, X*)-neighborhood of zero. Then there is a
incompact, convex, and balanced set S in X* such that 5o C ^. Since S is
^Abounded, it is also norm bounded in X* and there is a > 0 such that
S C *BX*. Then V D SQ D ±(BX*H = ±BX.
?
Lemma 4.37
Let E be a locally convex space with a countable local base. Assume that A
is a balanced convex set in E such that for every bounded set B C E there
is a > 0 satisfying ctB C A. Then A contains a neighborhood of zero in E.
Proof: Let {Un} be a countable local base for X. We may assume that
#n+i C Un for every n. By contradiction, assume that there is no n G N
for which Un C nA. Then there are Xfi
G Un such that xn ? nA. Clearly,
lim(zn) = 0. Since every convergent sequence is bounded, there is A' > 0
such that xn G KA for every n and thus xn G nA for n > I{. This is a
contradiction with xn ? nA for all n.
D
Proposition 4.38
Let X be a Banach space and r be a locally convex topology on X such that
(X, r)* = X*. If r has a countable local base, then r coincides with the
original norm topology of X.
Proof: By the Banach-Steinhaus theorem, the systems of all bounded
sets in (X, || • ||) and (X, w) coincide. By Corollary 4.35 and
Proposition 4.36, this system is the same for all locally convex topologies a for
which (X, a)* — X*. Applying Lemma 4.37 for the set A = Bx completes
the proof.
?
Representation and Compactness
Proposition 4.39 (Caratheodory; see, e.g., [Rud3])
Let D be a compact convex subset of an n-dimensional topological vector
space E. Every x G D is a convex combination of at most n -b 1 extreme
points of D.

122 4. Locally Convex Spaces
Proof: By induction on n. For n — 1, we have D = [a, 6], and the
statement follows. Assume that the result holds for n — 1. Consider D C E
with dim(E) = n and Int(D) ^ 0 (Exercise 4.18). If ? is a boundary point
of D, let H be a supporting hyperplane of D such that x G HDD. By
the induction hypothesis, a? is a convex combination of at most n extreme
points of HC\D which are extreme points of D (see the proof of the Krein-
Milman theorem). If x G Int(D), choose an extreme point zq G D and let
y be another boundary point of D that lies on the line going through x
and zq. By the same argument as above, y is a convex combination of at
n
most n extreme points z\,..., zn of D. Thus, y = ]P a2z2- with a; > 0 and
2 = 1
n
E <*i = i-
2=1
n
Now x = ay+(l — a)zo for some a G @,1), so x = 53 aa?-Zj + A — a)zo,
8 = 1
which is a convex combination of n + 1 extreme points zq, ..., zn.
?
We will now reformulate Caratheo dory's theorem in the language of
integral representation. Let D be a compact convex subset of an n-dimensional
topological vector space E. For y G D, let 6y be the Dirac measure at y.
Let x G -D be given. By Caratheodory's theorem, there are extreme
k
points a?i,...,?& of D, & < n + 1, and a2- > 0, 53 a^ — 1 such that
2 = 1
k k
x= ^ otiXi. Put // = 53 aifixi- Then /i is a probability measure on D. If
2=1 _ 2=1
/ is a continuous function on D, we have
If / is linear, we have f(x) = /E3 or,-a:,-) = 53 aif{xi) - Id f dljL-
This motivates the following definition.
Definition 4.40
Let D be a compact convex subset of a locally convex space E. We say that
a probability measure ji on D represents x G D if, for all f G E*,
/(*)= / fdp.
JD
If S is a Borel subset of D, we say that a measure ji is supported by S
i?n(D\S) = 0.
Proposition 4.41
Let D be a convex compact set in a locally convex space E. For every x G D
there is probability measure on D supported by Ext(D) that represents x.

4. Locally Convex Spaces 123
Proof: By the Krein-Milman theorem, conv(Ext(?))) = D. Hence there is
na
a net {ya} such that ya —> x and ya = Yl afxf^ where a? > 0, ^2 af = 1,
*=i
and x? E Ext(D). We represent each ya by a probability measure fia =
^af5x«. By the Riesz representation theorem, the set of all probability
measures on the compact set Ext(jD) is identified with a w*-compact set
in Bc,Ext,D<.y. Therefore, there is a subnet {t/^} of {ya} convergent in the
w*-topology to a probability measure /i on Ext(D). If / is a continuous
linear functional on ?", then its restriction to Ext(jD) is in C(Ext(D)) and
thus we have
/(*) = lim(/(W)) = lim J fdlip= J f
dfi.
Ext(D) Ext(?>)
To finish the proof, we extend // to ft on D by //(B) = //E D Ext(D))
for every Borel set B in D.
U
A natural question is whether a representing measure is supported by
Ext(-D) alone, as in C ar at heo dory's theorem. One of the difficulties is
that Ext(D) may not in general be a Borel set ([Phel]). However, if D
is metrizable, we have the following result.
Lemma 4.42
Let D be a compact convex set in a locally convex space E. If D is
metrizable, then the set of extreme points of D is a G$ set in D.
Proof: Let d be a metric on D that induces the topology of D. For n E N,
put
Fn = {xeD] x = ^,Vjze Ad(^) > ?}.
Then Fn is a closed set in D. Indeed, let Xk E Fn and lim(z&) — #, where
Xk = Vk+Zk for 2/jk, Zk E D, d(yk,Zk) > ^- Since D is a metrizable compact,
we may assume that lim(yjb) = y and lim(zfc) = z. Thus we have x = ^-.
Moreover, dOte,**) -+ d(t/,s), so d(y, *) > ~. This shows that ar E Fn,
proving that Fn is closed. To conclude the proof, observe that x E D is not
an extreme point of D if and only if x E Fn for some n.
?
Theorem 4.43 (Choquet representation theorem; see, e.g., [Phel])
Let D be a metrizable compact convex set in a locally convex space E. For
every xo E D, there is a probability measure /i on D supported by Ext(D)
that represents x$.
Proof: Note that Ext(D) is a Borel set by Lemma 4.42. Since D is a
metrizable compact, the space C(D) of all continuous functions on D with

124 4. Locally Convex Spaces
the supremum norm || • || is separable (Lemma 3.23). Let A denote the
subspace of C(D) formed by all continuous functions f on D such that
f(ax+/3y) — &f(x) + f3f(y) for every x,y G D and a,/? > 0 with a+/? = 1
and ax + f3y G D. Note that / G A need not, in general, be a restriction
onto D of some F(ar) = k + ip(x) for <p E E* (Exercise 4.21). Let Sa =
oo
{/ g A; ll/H = 1}, and let {hn} be a dense set in SA. Put h = ? 2~nh2n.
n = l
We claim that /i is a strictly convex continuous function on D\ that is,
that h(ax + A — <x)y) < ah(x) + A — a)h(y) whenever x,y ? D, x ^ y}
and a G @,1). Let x ^ y ? D be given. Choose n such that hn(x) ^ hn(y).
Using the affine property of hn and the fact that the real function t >-* t2
is strictly convex, for a G @,1) we have
h2n (ax + A - a)y) < a/?(s) + A - a)h2n(y).
Therefore,
ft(aa? + A - a)y) = 2-n/?(as + A - a)j/) + ?2-'A?(as + A - a)y)
< 2-nh2n(ax + (l-a)y)
+ J2^j^h](x)-^(l-a)h2(y))
< a2-nh2n(x) + (l-aJ-nhl(y)
= ah(x) + (l-a)h(y).
Let y be the subspace ofC(D) spanned by A and h (i.e., Y = A©RA).
For / G C(D), the concave envelope /: ?) —* R is defined by
f(x) = mi{g(x); geA,g>f}.
We list a few basic properties of /:
A) /is concave, bounded, upper semicontinuous, and thus Borel-measu-
rable;
B) / < /, and / = / if / is concave;
C) JTg < f + g; JTg = / + <7 if <7 G A; Ff = r / if r > 0;
D)|/-ff|<||/-d]
(to see D), write / = / - flr + flr < / - flr + ^ < ||/ - flr|| + flr).
Define a positively homogeneous subadditive functional p on C(D) by
P(g) — 9(xo), and define (p ?Y* by <p(g + rh) = g(x0) + rh(x0) for ^ G A
and r G R. Note that 9? is dominated by p on 7. We need only verify that
g(xo) + rh(xo) < g + rh(xo), which follows from C) if r > 0. If r < 0, then
g -\- rh = g -\- rh > g -{- rh by the concavity of # + r/i.
By the Hahn-Banach theorem, there is an extension m of cp onto the
whole C(D) such that ra(#) < g(xQ) for every # G C(D).

4. Locally Convex Spaces 125
If g E C(D) and g < 0, then 0 > g(x0) > m(g). Thus, if \\g\\ < 1, then
g — 1 < 0 and m(g) < m(l). Since m(l) = ^A) — 1? we obtain that m is a
norm-one functional on C{D). By Riesz's representation theorem, there is a
probability measure /ionD such that m(g) — fDg dfj, for every g E C(D).
Note that fDhdfi = h(xo) and <7(#0) = m(a?o) = f^gd^ for every # E A;
in particular, for every g E E*. To conclude the proof, it is enough to show
that:
(a) /^ Ad/i = fDh dfij
(b) {* E D\ h(x) = %)} C Ext(D).
Indeed, since h> h on D) from (a) we get ^{# E D; h(x) < h(x)} = 0.
By (b), it follows that D \ Ext(D) C {x E D; A(a:) > h(x)} and thus
/*(?> \ Ext(D)) = 0. To verify (a), write
I hdfj, < / /id/i < inf< / #d/i; g € A,g>h>
= inf J / # cfyz; # e A, g > A J = mf{#(zo); ? E A, # > h}
— Ji(xq) — cp(h) = m(/i) — I h dfi.
Jd
To prove (b), assume that x E .D\Ext(jD). Then a: = 2^, where y,z ? D,
y ^ z. Hence, using concavity of /i,
h(x) < \h{y) + ±A(z) < \h{y) + |%) < &(*±?) = %).
D
Theorem 4.44 (Banach, Dieudonne)
Zetf X be a Banach space, and let A be a convex set in X*. If AD nBx* is
w*-closed for every n E N; then A is w*-closed.
Proof: First, note that, under our assumption, A f] B is w*-closed for
every ball B in X* (centered at a general point of X*). Indeed, if n is large
enough so that B C nBx*, then A 0 B = (A n nBx*) H 5, where both 5
and A fl n??x* are iu*-closed sets.
Also note that A is norm closed. Indeed, if /n E -A, /n —> / in X*, then
/n is bounded and thus for some j we have that fn E jBx* for every n.
Since A. H iBx* is iu*-closed and fn E A D jBx* for every n, we have
Z EAnj?x* CA
We may assume without loss of generality that 0 E A. For n E N, put
An = Afl 2nBx*- Then A = (JAn and each An is w*-closed, convex,
and contains 0; therefore An — ((AnH)° by the bipolar theorem. Put G =
D(A)o.
We will show that A — G°, and this will finish the proof, because every
polar set is iu*-closed. First, observe that G C (An)o for every n and thus
G° D ((An)o)° = An for every n. Consequently, A = (J An C G°.
It remains to show that G° C A. This will be done in several steps.

126 4. Locally Convex Spaces
Claim A
(AnH C (An+iH + 2~nBx for every n G N.
Assume x ? (An+i)o + 2~nBx>By the separation theorem, we obtain
f € X* such that f(x) > 1 > sup{/(z); x G (An+iH + 2~nBx}. By the
definition of An, we get (An+iH D 2-(n+1)?x and /C • 2~(n^Bx) =
/B-(-+i)JBx + 2-x) C /((An+iH + 2"WB^). Therefore ||/|| < §2".
Also, sup{/(z); ar G (An+1H + 2"n5x} = sup (/) + sup (/) < 1
(An+iH 2-"-Bx
and thus sup (/) < 1 - 2-n||/||.
(An+i)o
Choose e G @,min(|,2-n||/||)) and put g = j^-J. From e < 2-n||/||
and sup (/) < 1 — 2~n||/||, we get sup (g) < 1; that is, g G
(An+i)o (An+iH
((An+i)o)° = An_j_i. In particular, g E A. Since ||<7|| < 2n (because
ll/ll < |2n and e < |) and ^ G A, we have # G An. Moreover, #(z) > 1
and thus ? ? (A-n)o- This proves Claim A.
Claim B
(AnH CG + 2-(n-^Bx for every n G N.
Given n G N and #n G (Ai)o> define inductively, using Claim A, vectors
arn+i, a?n+2, • • • in such a way that for m = n, n + 1,... we have xm+i G
(Am+i)o and \\xm — xm+i\\ < 2~~m. Then {xm} is norm Cauchy and thus
convergent to some x G X. Since (Am)o is norm closed for all ra, we have
00 1 1
that x G fl(^m)o = G. Moreover, \\xn - z|| < ? T^ZT < ^ZT-
Claim C
a= na+^-
Since A is convex and 0 E A, write a = 1x7A + e)a + (l — 1x7H for
ae A and e > 0 to see that Ac(l + e)A. Therefore Ac fl A + e)A
If X G H(l + 6)A an(^ ^ ~ (l + n)Xn W^n Xn ^ ^ f°r every n> tnen
a? — xn = ^xn and \\xn\\ < \\x\\. Thus #n —* #, and since A is closed, we
have a? G A- Claim C is proved.
Claim D
G°C f|(l + ^K
Let e > 0 and n G N be given. For every x and y in X,
« + y = (i + 0(if7 + ifef).
By this and Claim B, we have (AnH C A + e) conv{GU ^ij^x}-
Therefore, by taking polars, we have j^(G° n 2n~1eBx*) C An C A for every
n G N.

4. Locally Convex Spaces 127
Keeping e > 0 and taking the union over n, we have jx^G0 C A, which
means G° C A + e)A for every e > 0. Thus G° C f| (* +?:)^- Tnis Proves
Claim D, and Theorem 4.44 follows.
?
Let X be a Banach space. A set C C X* is called w*-sequentially closed
if {fn} CC}fn^f implies that / ? C.
Corollary 4.45
Let X be a separable Banach space, and let C be a convex set in X*. If C
is w*-sequentially closed, then it is w*-closed.
Proof: The ti;*-topology of Bx* is metrizable. In metrizable spaces,
the closures and sequential closures coincide, so since C fl nBx* is w*-
sequentially closed (in nBx*)) C fl nBx* is w*-closed in nBx* and thus
also in X* because nBx* itself is w*-closed.
?
Remark
If X — ?2 and xn — y/nen, where en is the standard unit vector in ^j
then 0 E {xn} (Exercise 3.15). Thus, the set {xn} is not w-closed in ?2
and yet {xn} CijBx is finite and thus weakly closed in X. Since on ?2
the w- and u>*-topologies coincide because ?2 is reflexive, we see that the
assumption of convexity of the set A in the Banach-Dieudonne theorem
cannot be dropped.
Corollary 4.46
Let X be a Banach space, and let F be a linear functional on X*. The
following are equivalent:
(i) F is w*-continuous.
(ii) F G X; that is, there is x ? X such that F(f) = f(x) for every f ? X*.
(Hi) The restriction of F to Bx* is w*-continuous.
(iv) F'1^) fl Bx* is w*-closed.
Proof: (i) is equivalent to (ii) by Theorem 3.16.
(i) => (iii) and (iii) => (iv) are trivial.
(iv) => (i): If (iv) holds, then by Theorem 4.44 and linearity of F,
jP_1@) is tt;*-closed. Hence i?~1(R\{0}) is u>*-open, so there is a iu*-open
convex neighborhood U of some / ? X* such that F ¦? 0 on U. We may
assume that F(f) < 0, then also F < 0 on U because U is connected and
F is continuous. Thus F is bounded above on some w*-neighborhood of
some point and hence w*-continuous on X* by Lemma 4.15.
?
Recall that a set A in a Banach space X is relatively weakly compact if
its closure in the weak topology of X is weakly compact. Note that a set
A C X is relatively weakly compact if and only if its w*-closure in X** is a

128 4. Locally Convex Spaces
bounded set in X. Indeed, then X equals A in A** and the latter one
is w*-compact in X** by the Alaoglu theorem.
Theorem 4.47 (Eberlein, Smulian; see, e.g., [Woj])
Let A be a subset of a Banach space and X. The following are equivalent:
(i) A is relatively weakly compact.
(ii) Every sequence {an} C A has a weakly convergent subsequence in X.
(Hi) Every sequence {an} C A has a weakly convergent subnet in X.
The last two conditions are called a relative weak sequential compactness
and a relative weak countable compactness, respectively.
Proof: (i) => (ii): (Smulian) Let {an} be a sequence in A. Denote V =
span{an}. Since it is separable, there is a sequence {fi} in X* that separates
points of V (Exercise 3.42). We can extract a subsequence {ank} such that
lim(/z(anfc)) exists for every i (c/., Exercise 3.111). We will denote this
subsequence by {an} again.
By (i), {an} has a cluster point in the w-topology of X. If a1, a2 are
arbitrary w-cluster points of {an}, then fi(al) — fi(a2) for every i. But
a1, a2 G V because V is unclosed; since {fi} separates the points of V, we
get a1 = a2. Hence {an} has a unique ^-cluster point a.
We will show that an —> a in X. Assume the contrary, so for some
fo € X* we can extract a subsequence {ank} such that lim(/o(anfc)) exists
and differs from /o(a). Let a1 be a ^-cluster point of {ank}. Then fo(ax) ^
/o(a), so a1 ^ a, which is a contradiction with the uniqueness of the cluster
point of {an}.
(ii) => (iii) is obvious.
(iii) => (i): (Whitley) By (iii), A is bounded in X. By the note above, it is
enough to prove that the iu*-closure of A in X** is contained in X. Assume
w*
the contrary; that is, there is some F E A \X. Let 8 = dist(F,X). We
will inductively construct a sequence {an} C A and a sequence {gn} C Bx*
such that:
A) F(gn) > f ? for n G N;
B) \gn(aj)\ < \S for j<n, j G N;
C) 9n{aj) > \8 for j >n,je N.
The construction proceeds as follows. Since ||F|| > <5, we choose gi that
w*
satisfies A). Since F is in A , we choose a\ such that |-F(^i) — <7i(ai)| is
small enough for C) to hold for j = I.
Suppose that ai,..., an and g±,..., gn have been chosen. By the Hahn-
Banach theorem, there is $ G X*** such that $(a2-) = 0 for i = 1,..., n,
$(F) > |<5, and ||$|| < 1. Thus, by Goldstine's theorem, there is gn+i G
w*
Bx* such that A) and B) hold. Next, we choose, using F G A and
an appropriate u>*-neighborhood, some an+i G A that approximates F on
<7i,..., gn+i so that C) is satisfied.

4. Locally Convex Spaces 129
We conclude the proof by showing that {an} has no ^-convergent subnet.
Assume that for some subnet {any} we have aUy —> a in X. By Theo-
M
rem 3.19, there is a convex combination ^ a&an^ with n7^ < nlN+1 ... <
77M such that
k=N
M
J2 akdnyh ~ a
k=N
<?•
By B), we have that if n > n7M, then \gn ( ? akan )
Nfc=iV
< ^<5. Therefore
|<7n(a)| < hS for some n > n7M. On the other hand, by C) we have gn(a) >
3 6 for every n. This contradiction completes the proof.
4
Let X — Co, A = Bx- Then A is not w-compact. Take F = A,1,...) G
4o, an = (l>...,l,0,...NA ^ = 1, and </n = @,..., 0,1, 0,...). Then
{an} and {gn} satisfy (l)-C) and in fact the proof of Theorem 4.47 can be
considered a generalization of this example.
Definition 4.48
A topological space T is said to have countable tightness if for every A C T
and x G A there is a countable set S C A such that x G S.
A topological space T is called angelic (or Frechet topological space) if for
every A C T and x G A there is a sequence {xn} C A such that xn —> x.
Example
Let T be uncountable. The element a = A,1,...) G ^oo(r) is in the closure
of 5Co(T) in the w*-topology by Goldstine's theorem. However, there is no
countable S C ?Co(r) sucn that a G S . Indeed, assume that a G S
for some countable S C ?Co(r)- Every element of cq(T) has a countable
support, and therefore there is a countable set M C T such that 5G) = 0
for every 7 (? M and every s G S. Then for a G S we have 0G) = 0
for 7 ^ M, a contradiction. Therefore, ^^(r) = ^*(r) m its w*-topology
does not have countable tightness.
Theorem 4.49 (Kaplansky; see, e.g., [Kot])
Every Banach space in its weak topology has countable tightness.
Proof: Let X be a Banach space, A C X, and x0 G A . Let k,m G N
be fixed. Given functionals /1,..., /& G #x* > we can choose 5 G ^4 fl {a: G
^5 |/i(^ — #o)| < ^jz — 1,...,&}. Since (s — #o) is w*-continuous on
X*, there are w*-neighborhoods Vi of /,- for i = l,...,fc such that for
g G Vi we have |#(s - x0)| < ^. For every fc-tuple /1,..., fk G ?**, find
s E A and then the neighborhoods V{ as before. Considering this for all
^-tuples fi,.. ., fk G Bx* , we obtain a family of neighborhoods of type
Vi x ... x Vfc. Since i?x* x ••• x Bx* is compact in its w*-topology, a
finite number of these products of neighborhoods cover Bx* x • • • x Bx* •

130 4. Locally Convex Spaces
Denote these neighborhoods by Vj x • • • x V?, I = 1,..., L. Collect the
corresponding sj. Then, given #1,..., gk G Bx*, there is sj G {si,..., sjo}
such that \gi(xo — Sj)\ < ^-.
Follow the same procedure for all fc, m in N to obtain a sequence Sk such
that given Ar-tuple /i,..., /jb G i?x* and j G N, there is s G {s&} such that
\fi(x0 ~ s)\< j for f = 1, 2,..., k. Therefore x0 G {sk} .
D
Example
Let xn = \/nen G ^2, where en is the standard unit vector in ?2- Then
— w
0 G {V^en} and there is no subsequence of xn that weakly converges to
0 (Exercise 3.15). Thus, ?2 in its weak topology is not an angelic space.
Theorem 4.50 (see, e.g., [Flo])
Every weakly compact set K in a Banach space X is angelic in its weak
topology.
Proof: Let A C K and x0 G ~^ C K. By Theorem 4.49, there is a
countable set S C A such that x0 G S . Put L = span(S). Then L is
separable. Furthermore, S C L O K and S is a weak compact set in a
separable space and thus metrizable in its weak topology (Proposition 3.29).
Therefore, there is a sequence {sn} C S such that lim(sn) = xq in the weak
topology of L, which means that lim(sn) = xo in the weak topology of X.
Since sn G A) the statement follows.
?
Recall that a subset A of a topological space X is called countably
compact if every sequence in A has a convergent subnet in A.
Theorem 4.51 (Eberlein, Smulian; see, e.g., [Woj])
Let A be a subset of a Banach space X. The following are equivalent:
(i) A is weakly compact.
(ii) A is weakly sequentially compact.
{Hi) A is weakly countably compact.
PROOF: According to preceding results, it is enough to show that A is
weakly closed whenever A is weakly sequentially closed and weakly
relatively compact. Let x G A . Since A is weakly compact, it is angelic in
the weak topology (Theorem 4.50) and thus there are xn G A such that
Xfi > X. Since A is weakly sequentially closed, we have x ? A.
n
Exercises
4.1 Show that if A is a balanced subset of a vector space V, then conv(v4)
is balanced.

4. Locally Convex Spaces 131
Hint: Use conv(A) = {Xa + A - AN; a, 6 G A, A G [0,1]}.
4.2 Let Lo be the vector space of all Lebesgue-measurable functions on
[0,1] with the metric p(x, y) = ? ^fe!^ *•
Show that p(xn,x) —> 0 if and only if xn —» x in measure, and that Lq is
a topological vector space.
Hint: Denote A? = {*; |(zn — z)(*)| > e}, jB? = {t\ \(xn — x)(i)\ < e}, then
f1 '*"-*' ,& = / '*"-*' ,*+/ ''»-«' ,*
70 l+|a?n-ar| J 1 + \xn - z| J 1 + \xn - x\
A. Bc
< ii{t]\(xn-x)(t)\>e}+e.
If xn does not converge to 0 in measure, assume fJ,(A?) > 8. Then
f1J^x\dt> I |«„-«| fzdt=jL
J0 l + \xn-x\ J l + \xn-x\ J l + e 1 + e
Ac Ac
and the latter does not converge to 0. We used here the monotonicity of
the function j^.
To observe that the convergence in measure generates a topology in which
both operations are continuous, write
|An#n ~ XqX0\ < |An| \xn - Xo\ + |An - Ao| \xo\
and note that
{t] |Anzn-A0zo| > z} C {t; |An| \xm-x0\ > e/2}U{t] |An-A0| |x0| > e/2}.
Measures of the sets on the right-hand side go to zero (for the second
summand use the Chebyshev inequality).
4.3 Let X be the vector space of all continuous functions on @,1). For
/ G X and r > 0, let V(f, r) = {g e X] \g(x)-f(x)\ < r for all x G @,1)}.
Let r be the topology on X that these sets generate. Show that the addition
is r-continuous but scalar multiplication is not.
Hint:{il}^0.
4.4 Let 0 < p < 1. Let ?p be the space of all sequences of real numbers x =
(x{) such that J2 \xi\p < °°> w^h the metric d(#, y) = Yl \xi ~ Vi\P• Show
that d is a complete metric and ?p is a topological vector space but not a
locally convex space. Show that ?* = i^ and the set U — {(#); X^ 1^*1 ^ 1}
has the property that f(U) is bounded for every / G ?p but U is unbounded
in ?p.
Hint: To show that ?* = ?00i note that the topology of ?p is stronger than
that of ?\ and thus every element of i^ is a continuous linear functional on
?p. If / ^ ^oo 5 we find x G ?p such that ^ f{X{ diverges. To see that ?p is not
locally convex, consider the set A = {xn} C ?p, where xn G ?p is denned by

132 4. Locally Convex Spaces
xn = np x if i — n and x™ — 0 otherwise. This is a relatively compact set
in ?p whose convex hull is not bounded since the sequence {yn} defined by
n
yn = n~l ^2 xk is unbounded in ?p. The latter is seen by an elementary
calculation, because <f@, y2n) can be estimated from below by
In A-pJ
{2n)~p ]T kp2~P > Bn)-pnnp2-p = ^— > oo.
Ar=n
4.5 Show that if {xn} is a sequence in a metrizable topological vector space
X and xn —> 0, then there are positive numbers 7n such that 7n —> oo and
InXn -* 0.
However, if X is the vector space of all real-valued functions on [0,1] with
the topology of pointwise convergence on [0,1], then there is a sequence
{fn} C X that pointwise converges to zero in X and if jn is any sequence
of numbers such that jn —> oo, then {jnfn} does not converge to zero.
Hint: The first part is standard, jn = . l For the second part, note
that the cardinality of the family of all sequences converging to zero is c.
4.6 Let X, Y be topological vector spaces. Assume that T is a linear map
from X onto Y such that T-1@) is closed in X and Y is finite-dimensional.
Show that T is a continuous and open map.
Hint: If U is a balanced neighborhood of zero in X and {e,-} is a basis for
y, then for some n,T(nU) D {e,-}. The convex hull of {±e2} contains 0
as an interior point. This shows that T is an open map. T is an algebraic
isomorphism of X/T~l{jS) and Y. Since all linear vector topologies on finite-
dimensional spaces coincide, and since the map x i—> x of X onto X/T_1@)
is continuous, T is continuous.
4.7 Let X be a normed space. Prove that X is complete in the canonical
metric given by its norm if and only if it is complete in its norm topology
as a topological vector space.
Hint: If every Cauchy net in X converges, then also every Cauchy sequence
(which is also a net) converges.
Assume that X is a complete normed space and {xa}aei is a Cauchy net
in X. Given n E N, find in G I, in > s'n-i, sucn ^na^ (x« ~ x/?) ? n^x ^or
all a,/? > in; in particular, za G B(x(n, ?) for all a > in. Then {#»„} is a
Cauchy sequence, which must converge to some x G X. We claim that x is
the limit of the net {xa}a^j: choose n G N. There exists no G N such that
no > n and ?2m G B(x, ?¦) for all m > no; in particular, ?z-n G 5(#, ^).
Then, for all a G I such that a > ino, we get xa G B(xin , ^-), and hence
za G B{xy ^). Consequently, a?a —> x.
4.8 Recall that y>: E1 —> F, where F, F are topological vector spaces, is
called a uniform homeomorphism of E and F if <?> is a bijection onto F

4. Locally Convex Spaces 133
and <p, <p~l are uniformly continuous. If such <p exists, the spaces E, F are
called uniformly homeomorphic.
Prove that if a locally convex space E is uniformly homeomorphic to a
Banach space X, then its topology is given by a norm (see [Bes2]).
Hint: Given a convex neighborhood V of 0 in E, find n G N such that
<p(x) — <p(y) G V whenever x — y G B^(^), where #x(n) *s tne °Pen ball
in X centered at 0 with radius ^. Note that by convexity of B^ (?) we have
Bx(^) + '" + Bx(^) = »*?(?) = B°(l) and thus, by the telescopic
argument,
V(B$) = ^(IJ + .-. + flO^))
C Vr + ,-- + 7 = n7.
Thus, <p(Bx) is a bounded neighborhood of E. E has a bounded
neighborhood of zero, so it is normable by Proposition 4.12.
4.9 Assume that fj G Vf(Q) and f(<p) — lim(/j(y>)) exists finite for every
3
(p G P(Q). Show that / G X>'(fi).
Hint: By the Banach-Steinhaus theorem, / is continuous on every T>k-
4.10 Let Q be an open set in Rn and {/;} C V'(Q) be such that {fj(<p)} is
bounded for every (p G V(Q). Show that there is a subsequence {fjk} and
/ G V'(Q) such that lim(/jfc(<p)) = /(y>) uniformly for <p on every bounded
subset of 2>@).
Hint: By the Banach-Steinhaus theorem, the restrictions of fj to ?># are
equicontinuous. Then use the Arzela-Ascoli theorem.
4.11 Let C°°[0,1] be the vector space of all real-valued functions on [0,1]
whose all derivatives are continuous on [0,1]. Let k G N and p G [l,oo).
Define the norm || • ||* on C°° by ||/||* = f ? ||/@||sV, where /«
yo<i<k '
denotes the /-th derivative of/ and || • || is the norm from Lp. The Sobolev
space Wp[0,1] is the completion of the normed space (C°°[0,1], || • || ). Show
that Wp[0t 1] is isomorphic to LP[0,1].
4.12 Prove directly that A0 is a if*-closed set.
Hint: A0 — f] {/; x(f) < 1} and the sets are w*-closed, since vectors from
X are iu*-continuous as elements of (X* ,w*)*.
4.13 Show that \J(Aa)° C (n^)°, (U^a)° = DO4*H, and (XA)° =
\AQ. Show that if Ax C A2, then (A2)° C (j4i)°.
Hint: Use the definition.
4.14 Show that (ByH = Bx*.

134 4. Locally Convex Spaces
Hint: Use the definition.
4.15 Let (X, || • ||) be a Banach space, and let Y be a closed subspace of
X*. For x e X, define ||z||y = sup{|j/*(a:)|; y E By}. Clearly, || • ||y is a
seminorm on X and || • || < || • ||. Show that:
(i) ||z||y = dist(z, Y1) (distance in X**).
v
(ii) Y is separating if and only if || • || is a norm on X if and only if
my1 = {0}inX**.
(iii) Y is norming if and only if || • || is an equivalent norm on X if and
only if dist(Sx, Yx) > 0 (distance in X**).
(iv) Y is 1-norming if and only if || • || = II * llx ^ an^ on^ ^
dist(Sx, Y1) = 1 (distance in X**).
Hint: (i): ||z||y = ||e||y* = ||a?||x**/yj- = dist(z, Y1). (ii) through (iv) are
routine.
4.16 Let X be a Banach space, and let Y be a separating closed subspace of
X*. Let a be the <x(X, Y)-topology on X (we know it is locally convex and
Hausdorff); denote BY = {x ? X; ||#||y < 1} (see the previous exercise).
Clearly, Bx C BY. Show that:
UBY = Bx<7: . ..—«,.
(ii) Y is norming if and only if Bx is bounded.
(iii) Y is 1-norming if and only if Bx is ^--closed if and only if the norm of
X is cr-lower semicontinuous.
Hint: (i): Consider (X, a); then_(X, a)* = Y. We have By = (Bx)° (polar
in Y) and BY — {{Bx)°)o — Bx • (ii) and the first two equivalences in
(iii) then follow using Exercise 4.15.The latter two conditions are equivalent
since aBx = || • ||" ((—oo, a]) for a > 0.
4.17 Assume that Bx is <r(X, Y)-closed for every norming subspace Y of
X*. Show that X is then reflexive.
Hint: By contradiction. Let G ? Sx** \ X be such that dist(G,X) < ~.
Put Y = G_1@). Then Y is norming (Exercise 3.43); hence Bx is <r(X, Y)-
closed and, by the previous exercise, Y is 1-norming. This is a contradiction:
since dist(G,X) < |, there is x ? X such that f(x) — G(f) < \ for all
/ € -#x*; in particular 11^| | > f. But then /(a:) <\< \\x\\ for all / ? ?y,
and Y cannot be 1-norming.
4.18 Let D be a convex set in a Rn. Show that either Int(Z)) / 0 or D lies
in a proper subspace of Rn.
Hint: Assume that OGi^, and let {e2}^:1 be the maximal linearly
independent set in D. Since D C span{e2}^:1, it remains to prove that lnt(D) / 0
if n = m. Consider C = conv{{ez}U {0}} C -D, and note that {^aze2;0 <
*i < ?} C C. Let z0 = E ^i- Then B = {X>,-et-; l<*i ~ 2^1 < 2^} is
an open neighborhood of xq and B C C C D.

4. Locally Convex Spaces 135
4.19 (Corson) Let A7bea compact convex set in a locally convex space E.
Show that if Ext(K) is countable, then K is metrizable.
Hint: Let X be a Banach space of all continuous affine functions on K with
the sup-norm. Identify K with its image in X*. By the bipolar theorem,
Bx* is the convex hull of K U (-K). Hence Ext(Bx*) is countable by
Theorem 3.41 and thus X* is separable by Corollary 3.50. Thus, X is
separable and hence (Bx*,w*) is metrizable, so K is metrizable.
4.20 Assume that C is a convex compact set in a locally convex space X,
and D is a closed subset of C. Show that if fi is a probability measure on
D, then there is a unique point x G C represented by /i. Moreover, the map
// >—> x is continuous from the w*-topology in C(D)* into the topology of
X. Prove this result.
Hint: For f e X*, put Hf = {x G D\ f(x) = //(/)}. We will show that
p| Hf DC ^ 0. Since C is compact, by the finite intersection property
fex*
it is enough to show that (p) Hfi) fl C ^ 0 for every finite set /i,..., /n G
X*. Define a map T from X into Rn by T{x) = (/i(*),..., /n(*)). Then
T(C) is compact and convex. It suffices to show that p G T(C), where
p = (f fi dfi,..., f fn dpi). Assume that p ? T(C). Then there is a linear
P D
functional on Rn that strictly separates p from the compact convex set
T(C) (i.e., there is a — (ai,...,an) such that (a,p) > sup{(a,T(#)); x G
C}, where (.,.) is the inner product on Rn). Define g G X* by g — Ylaifi-
Then we have
/ gdfi = / (]T a,-/i) d/i = ^ at- / /,- d/i = ]P a2//(/z) = (a,p)
> sup Y] ax-/i(aO = sup(flf).
c *^ c
This is a contradiction because /i(D) = 1.
If the map in question is not continuous as stated, then there are
measures fia and fi such that fia —> \x and their corresponding points xa
and x do not converge in X. Since C is compact, assume without loss
of generality that xa —* y / #. Then, for every / G X*, we have
f(xa) = fJ,a(f) —* fJ>(f) — f(x)- Because of continuity of /, we also have
f(y) = lim(/(#a)). Thus f{x) = f(y). Since X* separates points, we have
a contradiction.
4.21 Give an example of an affine function, in the sense as in the proof of
Choquet's theorem, defined on a metrizable compact subset D of a Banach
space X) that is not a restriction to D of a function of the form f{x) —
k + <p(x) for k GR, ^El*.
Hint: X = euD = {xeli', N < 4"f}, f(x) = ?2'zf.

136 4. Locally Convex Spaces
4.22 Let X be a separable Banach space. Assume that (Bx**,w*) is
angelic. Show that then every w*-compact convex set C in X* satisfies
C = conv(Ext(C)).
Hint: Use Choquet's theorem in (X,w); then the representation also holds
for / E X** by the Lebesgue dominated convergence theorem. Then use
the separation theorem. Note that it would follow directly from a general
version of Corollary 3.52.
4.23 Show that a Banach space X is reflexive if and only if the closed unit
ball of every equivalent norm on X* is uAclosed.
Hint: If the condition on w*-closedness holds and F E X** is given, consider
the set Vn = {/ E Bx*; |^(/)| < ?}• The Minkowski functional of Vn is an
equivalent norm on X*, and Vn is its closed unit ball. By our assumption,
Vn is w*-closed. Therefore, F~1(Q) H Bx* = f]Vn is u>*-closed and thus
F ? X, meaning that X is reflexive. If X is reflexive, then the closed unit
ball of any equivalent norm is lu-compact and thus w*-compact.
4.24 Let X be a Banach space. Show that if every separable subspace of
X is reflexive, then X is reflexive.
Hint: The Eberlein-Smulian theorem for the weak compactness of Bx-
4.25 (Klee) Let Ki,..., Kn be open convex sets in a locally convex space
X. Show that if f] K{ = 0, then there is a bounded linear operator L from
X into R"-1 such that f\L(Ki) = 0.
Hint: Let K — {(x\ - #2>?i - #3,..., a?i - xn)\ xi E K{,i = 1,.. .,n}.
Then K C Xn~l is an open convex set that does not contain 0. Thus,
by the Hahn-Banach theorem, there exist x\, x\)-.., xn-\ E X* such
n-l
that Yl xi(xi ~ xi) > 0 f°r a^ xi ? K(> i — 1}...,^. Put L(z) =
s = l
(**(*),..., <_i(s)) for x E X.

5
Structure of Banach Spaces
Definition 5.1
Let X be a vector space, A linear map P:X —> X is called a projection
onto a subspace Y of X if P(X) = Y and P(y) = y for every y ? Y'.
Let Mi, M.2 be subspaces of a vector space V such that V = Mi + M2.
Recall that it is a tfrreci sum (write 7 = Mi 0 M2) if Mi n M2 = {0}.
Equivalently, for every v & V there is a unique pair (^1,^2) G Mi x M2
such that v = vi -\- V2- By the uniqueness of the decomposition, the maps
Pi(v) — V{ for i — 1,2 are well defined. We observe that Pf.V —> V are
projections onto M,- for i = 1,2 and Pi + P2 = Iv •
Given a subspace Mi of a vector space V, there is another subspace
M2 such that V = Mi 0 M2. Such a subspace M2 is called an algebraic
complement of Mi in V.
Definition 5.2
A subspace Y of a Banach space X is said to be complemented in X if
there is a bounded linear projection of X onto Y.
Let Y\ be a closed subspace of X. We say that Y2 is a (topological)
complement of Yi in X if X — Y\ 0 Y2 and Y2 is a closed subspace of X.
Note that if Y is a complemented subspace of a Banach space X, then Y is
closed. On the other hand, not every closed subspace of X is complemented.
However, finite-dimensional and finite-codimensional closed subspaces are
complemented (Exercise 5.24).

138 5. Structure of Banach Spaces
Proposition 5.3
Let Y be a closed subspace of a Banach space X. Y is complemented in X
if and only if there exists a topological complement ofY in X.
Proof: Assume that there is a closed subspace Z of X such that X =
Y 0 Z. Let P be the linear projection onto Y such that Z — Ker(P). We
will show that the graph Q — {(#, P(x))] x E X} is closed.
Let xn E X be such that (xn,P(xn)) —* (#, y). Since Y is closed, y EY.
Note that (xn - P(xnj) E Z = Ker(P). Indeed, P(xn - P(xn)) = P(xn) -
P2[xn) = 0. Since Z is closed, we get x — y E Z. By the uniqueness of the
decomposition, we get y — P(x). Thus (x, y) E Q, and by the closed graph
theorem, P is a bounded projection onto Y.
To prove the opposite implication, assume P2 — P) P(X) — Y. We will
show that Ker(P) is a complement of Y.
lixe Ker(P)nP(X), then x = P(x) = 0, so YflKer(P) = {0}. Consider
any x E X. Since P is a projection, x — P(#) E Ker(P), and we get a
decomposition x = P(#) -f (a; — P(#)). This shows that X — Z ^ Ker(P).
Since P is continuous, Ker(P) is a closed subspace of X.
U
Observe that if H is a Hilbert space and F its closed subspace, then the
orthogonal complement F1 of F is a complement of F in the sense of the
above definition, and the orthogonal projection of H to F is a norm-one
projection onto F in the sense of Definition 5.1.
Fact 5.4
Let X, Y be Banach spaces, let T be an isomorphism of X onto Y. If X\ is
a complemented subspace of X with topological complement X2, then T(Xi)
is a complemented subspace ofY with topological complement T(X2).
PROOF: Clearly, Y = T(Xi) 0 T(X2) and both subspaces are closed.
?
Note that if P is a projection of X onto X\ with Ker(P) = X2, then
Q = TPT~l is a projection ofY onto T(Xl) with Ker(Q) = T(X2).
Let X be a normed space such that X = Y 0 Z for complemented
subspaces Y, Z. Then. X is. isomorphic to the direct sum (Y 0 Z)^ of
spaces Y,Z with the norm ||(y, z)|| = max(||i/||, ||^||).
Proposition 5.5
Let Y be a closed subspace of a Banach space X. IfY is complemented and
Z is a complement ofY in X} then.X/Y is isomorphic to Z.
The dual X* is then isomorphic to Y* 0 Z*; in short, (Y 0 Z)* =Y* 0 Z*.
// Y is a closed subspace of a Hilbert space H and Z is the orthogonal
complement ofY in H, then H/Y is isometric to Z.
Proof: Consider an operator I:Z -> X/Y defined by I(z) - z E XjY.
From the definition of the norm of X/Y, it follows that 7 is a continuous

5. Structure of Banach Spaces 139
linear operator. We claim that / is one-to-one. To see this, assume that for
z G Z we have z = 6. Then z ez = 0 = Y. Therefore, z G Z n Y and thus
z — 0. I is also an onto map. Indeed, let x G X/Y. We must show that
there is x G x such that # G Z. Pick arbitrary z G x. Write x = y + 2,
where y G Y, z G Z. Since y G Y, we have z = # + (—y) G ?. Therefore, /
is an isomorphism of Z onto X/Y by the open mapping theorem.
Let P, resp. Q, be projections of X onto Y, resp. Z. By Exercise 5.10,
Y* = P*(X*) and Z* = Q*(X*) are complemented in X\ Moreover, from
Ix = P + Qwe get Ix* = P* + <9*; that is, Y* + Z* = X*. It remains to
show that Y* fl Z* = {0}. But we have Ker(P) = Q(X), and hence
Q*(X*) C Ker(QI = P{X)L = Ker(P*).
Thus if / G <2*(X*) H P*(X*), then / G Ker(P*) and / = P*(/) = 0.
If H is a Hilbert space, Y is a closed subspace of P, and Z = YL, then
consider the map / as before. If z G Z, then dist(z, Y) = ||z|| by the proof
of Theorem 1.33. Therefore, J is an isometry.
?
Recall the definition of the Kronecker delta: Sij = 0 if i / j and 8(j = 1
iff = j.
Let X be a Banach space, and consider vectors {ei,..., en} C X. Func-
tionals {/1,..., fn} C X* are called biorthogonal to {e;} if /2(ej) = 6{j for
2,y = 1,..., n. The set {e2-; /j} is then called a biorthogonal system in X.
A biorthogonal system {e2-; /;} is called an Auerbach basis of X if {e;}"-!
is a basis of X and ||et-|| = ||/2|| = 1 for every i.
Theorem 5.6 (Auerbach; see, e.g., [LiT2])
(i) Let X be a Banach space. If dim(X) = n; tfAerc /Aere exists an Auerbach
basis {e;;/J?=1 ofX.
(ii) Let Y be a subspace of a Banach space X. //dim(Y) = n, then there
exists a projection P of X onto Y such that \\P\\ < n.
Proof: (i): Let {#i,..., xn} be an algebraic basis of X. For t/i,..., un G
Bx let v(ui,..., un) be the determinant of the matrix whose j-th row is
formed by the coordinates of Uj in the basis {#i,..., xn}. The function \v\
is continuous on the compact set Bx x • • • x Bx, and therefore there is
(ei,..., en) G Bx x • • ¦ x Bx such that
v(ei,...,e„) = max{|u(zi,...,xn)|; (a?i,.. .,a?„) G P* x ... x Bx}-
Since determinants are homogeneous in each coordinate, we have e2- G Sx-
Because v(ei,...,en) ^ 0, the vectors {e;} are linearly independent, so
they form a basis of X. For i = 1,..., n, define /2- G ^* by
v(ei,...,et-_i,ar,et-+i,...,e„)
/»W = 7 x •
v(ei,e2,...,en)

140 5. Structure of Banach Spaces
Then fi(ej) = 6ij, so {{ei,..., en}; {/l3..., /n}} is a biorthogonal system.
Moreover, sup{|/j(#)|; x E Bx} < 1 for each j, and therefore ||/j|| = 1.
Thus, {ei]fi} is an Auerbach basis of X.
(ii): Let {e2;/2} be an Auerbach basis of Y. We extend /; to norm-
one functional on X. Then we define an operator P:X —*• Y by P(x) =
n
13 fi(x)ei fox x e X.
*=i
We claim that P is a projection onto Y. Indeed, for every y = ]T} aiei ?
n
Y, we have a2- = /,-(?/). Therefore, P(y) = J2 fi(y)ti = V- Finally, if x G X
and ||e|| < 1, then ||P(x)|| < ? |/f(x)| ||e,-|| < ? 1 = n.
2=1 t=l
n
Theorem 5.7 (Lindenstrauss, Tzafriri; see, e.g., [T-J])
Let X be a Banach space. Every closed subspace of X is complemented in
X if and only if X is isomorphic to a Hilbert space.
For the proof of the "only if" implication, we refer the reader to [T-J].
The other direction is easy: If T is an isomorphism of X onto a Hilbert
space H, and Y is a closed subspace of X, then T{Y) is closed and hence
complemented by some projection P in H, and T~XPT is a projection of
X onto Y.
The next result—and the method of its proof—are part of the so-called
PelczynskVs decomposition method.
Proposition 5.8
Let X and Y be Banach spaces. Assume that X is isomorphic to X 0 X,
Y is isomorphic to Y (BY, X is isomorphic to a complemented subspace
of Y, and Y is isomorphic to a complemented subspace of X. Then X is
isomorphic to Y.
Proof: We will write X ~ Y to mean that the spaces X, Y are isomorphic.
We have X ~ Y 0 Z for some Z and Y ~ X 0 W for some W. Then
107 - Y®Z®Y - Y0Z - X. Similarly, we have X0Y - X®X®W -
X 0 W - Y. Therefore X~Y.~
?
We note that Gowers constructed Banach spaces X, Y such that X
is isomorphic to a complemented subspace of Y, Y is isomorphic to a
complemented subspace of X, and X is not isomorphic to Y ([Gow3]).
Theorem 5.9
Every separable Banach space is isometric to a quotient space of l\.
PROOF: Let X be a separable Banach space. Let {y1} be a dense sequence
00
in Bx • Define a map T from l\ into X by T(x) — J2 xiVl f°r x — (x%) ? ^i •
2 = 1

5. Structure of Banach Spaces 141
First, we note that, for every x G ii, the series J2 xiy1 is absolutely
2 = 1
OO
convergent since J2 Ik^ll < Yl \xi\ — \\x\h < °°- The map T is thus a
2 = 1
bounded linear operator from t\ into X and ||T|| < 1.
Define the map f:?i/Kei(T) -+ X by f (x) = T(x), where ? is a coset
in <?i/Ker(T). Given \\x\\ < 1, there is x G ? with ||z|| < 1 and hence
||f (x)|| = \\T(x)\\ < 1. Thus, we have f (B°/Ker(T)) C 5°.
By 5° C T(B°) and Lemma 2.23, B° C T(B°) C f (B°/Ker(r));
hence T(B° ,K,TC) = 5^. Since T is one-to-one, it is an isometry
(Exercise 1.26).
?
Similarly, we can prove that, given a Banach space X, there is a set T
such that X is isometric to a quotient of ^i(r).
Note that if X is not isomorphic to l\ and T\, T2 are bounded linear
operators from l\ onto X, then Ker(Ti) is isomorphic to Ker(T2) ([LiRl],
see also [LiTl]).
We say that a Banach space X has the lifting property if for all Banach
spaces Y, Z such that there is an onto operator S G B(Y, Z) and for all
T G B(X, Z) there is f G fl(X, Y) such that T = Sof.
In particular, if a Banach space X has the lifting property and is
isomorphic to a quotient of Y, then it is isomorphic to a subspace of Y. Indeed,
it follows from the lifting property for Z — X, T — Jj, and S being the
canonical quotient map.
Proposition 5.10
?1 has the lifting property.
Proof: Let xi = T(e;) for i G N, where a denote the standard unit
vectors of l\. By the open mapping theorem, there is a constant C such
that for every i there is yi G Y with S(yi) — X{ and ||t/;|| < C. We define
f (et-) = Vi. Then 5(f (ef)) = S(yi) = T(et-), and by linearity, 5 o f = T.
From jB^ = conv{ez} we then get ||T|| < C.
?
Proposition 5.11
Every separable Banach space is isometric to a subspace of l^.
PROOF: By Proposition 3.22, there is a sequence {/$} that is w*-dense in
Bx*. Define T:X -> 4o by T(z) = (/*(*)),.. Then \\x\\ = sup|/f-(a?)| =
11^(^)|100 and T is a linear isometry into t^.
?

142 5. Structure of Banach Spaces
Similarly, we prove that every Banach space is isometric to a subspace of
^oo(r) for some I\ Note (Exercise 5.49) that a dual space to an arbitrary
separable Banach space is also isometric to a subspace of t^.
Proposition 5.12
Every separable Banach space is isometric to a subspace o/^oo/co-
Proof: Let {x*} be a tfAdense sequence in Bx* • Define T: X —> ?& by
T(x) = (x{(x), x*2{x), x\[x), xl(x), xl(x), x\{x),...).
As in Proposition 5.11, we prove that T is a linear isometry into i^. Let
q : ioo —* ^oo/co be the canonical quotient linear operator. We claim that
q o T is an isometry from X into ^oo/co- This follows from the fact that
\\q{xi)\\ = HOc.Olko/co = limsup \x{\ (Exercise 1.20).
D
Recall that a Banach space X is called injective if for every Banach space
Y such that X is a closed subspace of Y there is a projection of Y onto
X. In other words, X is complemented in all superspaces. The following
theorem shows that 4>o actually has a slightly stronger property.
Proposition 5.13
Let Y be a subspace of a Banach space X.IfY is isomorphic to t^, then
Y is complemented in X.
IfY is isometric to l^, then there is a norm-one projection of X onto Y.
Proof: Let T be an isomorphism of Y onto l^, T(x) = (T(x)i) for x E Y.
For i e N, define /,- E Y* by f{(x) = T{x)i\ note that ||/t-|| < ||T||.
We extend fi to functionals on X of equal norm. Define Q:X —> i^ by
Q(x) = (fi(x)) and P:X -* Y by P(x) = T-l(Q(x)). Then ||Q|| < ||T||
and hence ||P|| < \\T-l\\ \\T\\. We claim that P is a projection onto Y.
Let y E Y. Note that <2(y) = T(y); hence we have P(y) = T~1Q(y) =
n
Theorem 5.14 (Sobczyk; see, e.g., [LiT2])
Ze^ 7 Se a closed subspace of a separable Banach space X. If Y is
isomorphic to cq, then Y is complemented in X.
Proof: (Veech) Let T be an isomorphism of Y onto c0; we may assume
that ||T|| = 1. For x E Y, we have T(x) = (T(x)i) E Co, so we can define
fi e Y* by fi(x) = T(x)i. Note that \\fi\\ < \\T\\ = 1 and extend /,- to a
functional on X of equal norm. Denote F = Bx* HY1 and consider the
metric p that induces the u>*-topology on Bx* (see Proposition 3.22).
We claim that limp(/n,F) — 0. Indeed, let {fnk} be a subsequence
n
such that limp(fnkiF) = limsupp(/n,F) — a > 0. Since Bx* is w*-
n n
compact by the Alaoglu theorem, and the it>*-topology is metrizable, we

5. Structure of Banach Spaces 143
may assume that there is / E Bx* such that fnk —» /. We observe that
T{x)n —> 0 for every x E Y, and hence also fn(x) —> 0. Thus, f(y) —
lim(/nfcB/)) = 0 for all yeY. Therefore f\y = 0, and hence f e F. This
shows that a < limsupn p(fnk, /) = 0 (because p is i/;*-continuous on Bx*)
and consequently a = 0.
Choose gn e F so that p(fntgn) < p(fn,F) + ?. Since p(fn,F) -+ 0,
we get p(fnjgn) -+ 0, so fn ~ 9n ^ 0. Thus, or h+ ((/,- - 0x)O*O)ieN is a
bounded and linear operator from X into Co; call it Q.
Define a bounded linear operator P\X —» Y by P(a?) = T (Q(z)). We
conclude the proof by showing that P is a projection onto Y. Choose any
y E Y. Note that gn(y) = 0, and hence Q(y) = (fi{y))ieN = (T(y)i)ieN =
T(y). Thus, P(y) = T~lQ(y) = T~lT(y) = j/ as required.
?
Note that if a separable Banach space # is complemented in every
separable overspace X, then Z is isomorphic to cq (Zippin [Zip]).
Examples may be found in [PeBe] of non-complemented subspaces of ?p
that are isomorphic to ?p. On the other hand, the subspace span{e2;} is
complemented in and isometric to ?p.
Theorem 5.15 (Phillips [Phi])
The space Co is not complemented in l^.
For the complementability of cq(T) in overspaces, see for example [GKL].
In the proof of Theorem 5.15, we will use the following lemma.
Lemma 5.16 (Sierpiriski)
There exists a family T of cardinality continuum formed by infinite subsets
ofN such that ifFi,F2 E T, Fx ^ F2, then card(Fi n F2) < oo.
Proof: Let i: N —> Q be a one-to-one and onto map. For every real number
r, choose a sequence of distinct rationals {#?} convergent to r. Define a map
^:R^NNby^(r) = {.-1(^)}r=i-
For ri,r2 G R, n ^ r2j we have ca,id({i-1(grn1)}neNn{i-1(grn2)}neN) <
oo. Putting T = {(p(r); r E R} concludes the proof of this lemma.
?
Proof of Theorem 5.15: (see, e.g., [Jan]) Assume, by contradiction, that
P is a bounded linear projection of i^ onto Co, and put Q — I — P, where
I is the identity map of i^. For j E R, put xy = XFy (the characteristic
function of F1 in N), where {F7}7gR is the family constructed above.
Claim
For every e > 0 and n E N, we Aaue card{7 E R; |Q(#7)n| > ?} < oo.
Indeed, assume that |Q(#7i)n| > s for z — 1,..., k. Define

144 5. Structure of Banach Spaces
Then x^
-x'7l = X
(Fyin\J{Fyj', j^,i=i,..,fc})
i
1:
k
E cq because the intersection
is finite. Thus we obtain Q(xli) = Q(x'1%) for i — \
k
.., k. We have
Q(^sign[Q(*;.)n]*;.)n = ? sign [<?(*;.)„]<?(*;.)„
2 = 1
k k
2 = 1
On the other hand,
supports of xly., z = 1,.
i=i
k
?sign[Q(a^.)n]a^.
2 = 1
. , ?. We thus have
2 = 1
< 1 due to the disjoint
ke < Q(^8ign[Q«.)n]x;.)w
2 = 1
k
< ||Q|||^sign[(Q4JJ4t
Therefore k < ^.
2 = 1
<IIQII-
Using the claim, we see that A— (J {7 E R; Q(#7)n / 0} is countable.
n=l
Therefore, there is 7 E R \ A for which we have Q{x1)n — 0 for every
n. Hence Q(x7) = 0, and thus x1 — P(x1). This implies that #7 E Co, a
contradiction with the fact that x<y is a characteristic function of an infinite
set.
?
In fact, we will see in Chapter 6 that cq is not even isomorphic to a
quotient of l^. However, there exists a Lipschitz map from i^ onto cq
whose restriction to cq is the identity ([Linl], Exercise 5.43).
Theorem 5.17 (Banach, Mazur)
Every separable Banach space is isometric to a subspace of C[0,1].
Proof: Let X be a separable Banach space. Then Bx* is a compact metric
space in the w*-topology; therefore, there exists a continuous map 9? from
the Cantor set D C [0,1] onto (Bx*, w*) (see, e.g., [Wil]). Define a map T
from X into G(D), the space of continuous functions on D, by T(x): d t-»
((p(d))(x). Then T(x) is a continuous function on Z), and T is a well-defined
map. Now we define a map T from X into C[0,1]. Given x E X, extend
the function T(x)(d) defined on D onto [0,1] as follows. For r ? D, put
T(x)(r) = aT(x)(ri) -f- /3T(x)(r2), where r\ — max{p E D] p < r}, r2 =
min{p GD; r < p}, and r = ar*i + /?r2-

5. Structure of Banach Spaces 145
T is a linear map from X into C[0,1] and
sup \T(x)\ = sup \f (x)\ = sup |^(d)(a?)| = sup |/(a?)| = ||x||.
[0,1] d?D deD f?Bx*
Therefore, T is a linear isometry from X into C[0,1].
D
We say that C[0,1] is isometrically universal for all separable Banach
spaces. Note that there is no finite-dimensional Banach space isometrically
universal for all two-dimensional Banach spaces ([Besl]).
It is conjectured that every complemented subspace of C[0,1] is
isomorphic to C[0,1] or to C{K) for some countable metric space K.
Corollary 5.18 (Banach, Frechet, Mazur)
Every separable metric space (Pyp) is isometric to a subset o/C[0,1].
Proof: Let {xn}nK>=1 be a dense sequence in (P,p). Choose xq G P and
define a map cp from P into l^ by <p(x) = (p(x, xn) — p(xoixn)) .
Note that, given x G P, by the triangle inequality
\p(x, xn) - p(x0jxn)\ < p{x, a?0)
and therefore indeed (p(x) G ^oo • We now show that (p is an isometry from
P into ^oo- First, if x and x' are in P, then
\\<p(x) - p(z')lko = SUP \p(x> xn) ~ p{x', Xn)\ < p{x, X!).
n
We claim that ||^(a?) — y>(?7)lk» ^ p{xix') for every #,#' in P. Given e > 0,
by density choose n G N such that p(#, a?n) < e/2. Then
p(x', xn) > p{x\ x) - p(x, xn) > p(x', x) - e/2.
Therefore
\\(p(x) - <p{x')\\ioo > M*'* Xn) ~ p(x, xn)\ > p(x', Xn) - p(x, Xn)
> p(x\ x) - e/2 - e/2 = p(xf, x) - e.
Thus, \\<p(x) — <?>(?')Ikoo = p(x'> x) because e > 0 was arbitrary.
We see that <p is an isometric bijection of P onto a separable subset
<p(P) of -?oo. Let X = span (y>(P)) in ^oo. Then X is separable and there
is a linear isometry T of X onto a separable subspace of C[0,1]. Then the
composition T o ^ is an isometric bijection of P onto a separable subset of
C[0,1].
?
Recall that a sequence {xn} C X is weakly Cauchy if {/(#n)}^i is
Cauchy (that is, convergent) for every f ? X*. Note that, in general, a
weakly Cauchy sequence in a Banach space need not be weakly convergent.

146 5. Structure of Banach Spaces
Theorem 5.19 (Schur)
Let {xn} be a sequence in l\. If {xn} is weakly Cauchy, then {xn} is norm-
convergent in t\.
Proof: First, we show that if xn —> 0 in ?i, then xn —» 0. By contradiction,
assume that for some e > 0 there is an increasing sequence n\ < ri2 < ...
oo oo
such that ||arn;|| = ? \XV\ > e- Choose Nx so that ? {x1?1] < |.
z = l i=Nx+l
Then ? |^^| > §?, which is equivalent to ? e^x?1 > §?, where ?2ni =
2 = 1 2 = 1
sign (a?^1) for i = 1,..., iVi. Note that if we choose an arbitrary sequence
of signs {si — ±1} such that Si — e™1 for i < Ni, we have
oo Ni oo iVi oo
|E^?i| = |x>?i*?i + Yl ^|>|X>r*r|- E n
2 = 1 2 = 1 *=JVi+l 2 = 1 2=JVi+l
Next, find nj2 such that J2 \xi32\ < |- This is possible since eJ1 —> 0.
2 = 1
OO .
Then we choose A^2 > Ni such that ^ |z2-J21 < | and consequently
i=N2+l
N* n
J2 \xiJ2\ ^ te' Then, for arbitrary choice of signs {si — ±1} satisfying
2 = 1
Si = e7}1 for i < Ni and a = ??2 for Ni <i < N2) we have
oo iV2 JVi oo
|E^i > | E ««?*|-X>?*i- E i*?*i
2 = 1 2=iVi+l 2 = 1 2=iV2 + l
N2 N2 JVX
> I E ««r* -f^XX* •-!>"* >-!<
2 = ATi-fl 2 = 1 2 = 1
— 5C 5 5C 5*
Repeating this process, we obtain a vector u = (u2) E ^oo defined by
Kf = ??fe for iVfc_i < i < NjZ) which has the property that u(;rnj*) > | for
all fc. This is a contradiction with xn —» 0.
For a weakly Cauchy sequence, we proceed in an analogous way: If {xn}
is weakly Cauchy and not norm Cauchy, there exists e > 0 and indices
njj mi ~~* °° sucn that lknj "" xTnj\\ > ?• Then consider the sequence
{arnJ — #mj'}j, which converges weakly to zero.
?
We say that a Banach space X has the Schur property if every weakly
convergent sequence is norm-convergent. We have just shown that l\ has
this property.

5. Structure of Banach Spaces 147
Exercises
5.1 (i) Let M be a subspace of a vector space V. Show that there is a linear
projection of V onto M (i.e., P\ = Im and P(V) = M).
(ii) Show that if a linear map P: V -* V satisfies P2 = P, then 7 = P(V)®
Ker(P). Moreover, Q = ly — P is a projection such that Q(V) = Ker(P)
and Ker(Q) = P(V).
Hint: (i): Extend an algebraic basis {xa} of M to an algebraic basis {#a, yp}
of V and define P by P(xa) = xa, P{yp) = 0.
(ii): Any x EV can be written as x — P(x) + (x — P(x)).
5.2 Let P be a bounded linear projection in a Banach space X. Show that
P* is a projection in X*.
Hint: (P*J = (P2)* = P*.
5.3 Let P, Q be projections in a Banach space X. Show that the following
are equivalent:
(i) P(X) C Q(X) and P*(X*) C Q*(X*).
(ii) PQ = QP = P.
(iii) P(X) C Q(X) and Ker(Q) C Ker(P).
Hint: P(X) C Q(X) iff QP = P. Using dual projections, (i) and (ii) are
therefore equivalent. Assume (ii) holds, and let Q(x) — 0. Then P(x) =
PQ(x) = P@) = 0. If (iii) holds, then QP = P. Moreover, given x E X,
PQ(z) - P(z) = P(Q(a?) -x)=0 since Q(Q(ar) - x) = 0. Thus (ii) holds.
5.4 Let P be a bounded linear projection of a Banach space X onto P(X).
Show that, for every x G X,
dist(a?, P(X)) < ||* - P(x)|| < (||P|| + 1) dist(x, P(X)).
Hint: Obviously, ||a? - P(a?)|| > dist(x, P(X)). On the other hand, if y G
P(X), write a: - P(x) = x-y + y- P(x) = x - y + P(y) - P(x). Thus,
II* - p(*)|| < Ik - y\\ + \\p(*) - P(y)\\ < (i + ||P||)lk - y||.
5.5 Let Y be a subspace of a Banach space X. Show that if there is a
bounded projection P onto Y, then Y is closed.
Hint: Take yn G Y such that yn —> y. Then P(yn) —> ^(y)? so also 2/n —>
P(y). By the uniqueness of limit, y = P(y) and y EY.
5.6 Assume that Y{ are subspaces of a Banach space X such that X =
^i 0 Y2 (algebraic sum). Let Pi be the associated linear projections onto
Yi (so Px + P2 = Ix). Show that:
(i) PX(X) = Ker(P2).
(ii) Pi is bounded if and only if P2 is bounded.
(iii) Both Yi and Y2 are closed if and only if both Pi are bounded.

148 5. Structure of Banach Spaces
Find an example when Y\ is closed but Y<i is not.
The complement of a subspace Y is sometimes defined as a subspace Z
such that X = Y 0 Z and the corresponding projections are bounded, (iii)
shows that it is an equivalent definition, and the previous exercise shows
that closedness of Y is necessary for the existence of a complement.
Hint: (i): x E Ker(P2) satisfies x = (Pi + P2)(x) = Pi(#), and similarly the
other inclusion, (ii): P2 = Ix — Pi- (iii): Proposition 5.3.
For the example, Y\ — R, I2 = Ker(/) for some discontinuous linear
functional /.
5.7 Let X be a Banach space. Show that X* is complemented in X***.
Hint: Define P:X*** -» X* by P(f) = /|^. This is called Dixmier's
projection; note that ||P|| = 1.
5.8 Let Y be a complemented subspace of a Banach space X. Let P be
a projection of X onto Y. Show that the dual operator P* is a map that
extends elements of Y* to elements in X*. If ||P|| = 1, we get a linear
Hahn-Banach extension.
Let Y be a closed subspace of a reflexive Banach space X. Assume that
there is a bounded linear operator E: Y* —> X* such that E(f) is an
extension of / on X with the same norm. Show that Y is then complemented
in X.
Hint: Show that E* is a projection of X onto Y.
5.9 Let X be a Banach space. Show that if Y is a complemented subspace
of X (by a projection P), then Ker(PI is a complemented subspace of X*
(by the projection P*) and Y1 = Ker(P*) is a complemented subspace of
X\
Hint: Clearly, P*(X*) C Ker(PI. On the other hand, fix y* e Ker(PI.
For x G X, write z = t/ + 2 with y EY and z G Ker(P), then P*(y*)(x) =
t/*(z), so P*(y*) = y*. Ker(P*) = YL is straightforward.
5.10 Let X be a Banach space, and let P be a bounded linear projection
of X onto P(X). Show that P(X)* is isomorphic to P*(X*).
Hint: Note that P* can be considered a map from P(X)* into X*. Check
this operator for the isomorphism.
5.11 Show that if Y is isomorphic to a complemented subspace of X, then
Y* is isomorphic to a complemented subspace of X*.
Hint: Let Y be isomorphic to Z, and let P be a projection of X onto Z.
By Exercise 5.10, P*(Z*) is closed in X* and isomorphic to Z*, and hence
to Y*. By Exercise 5.2, P*(Z*) is complemented in X*.

5. Structure of Banach Spaces 149
5.12 Let X be a normed space, and let Y, Z be subspaces of X such that
X = Y 0 Z. Assume that there is a bounded linear projection P of X onto
Y with Ker(P) = Z. Show that:
(i) X is isomorphic to (Y 0 Z)^ (hence to (Y 0 Z)p for p > 1).
(ii) If Y, Z are complete spaces, then so is X.
Hint: (i): Define T(yy z) = y + z. Since X = Y + Z<mdYnZ = {0}, this
is a bijection. We have \\T(y,z)\\ = \\y + z\\ < \\y\\ + \\z\\ < 2max(||y||, \\z\\).
On the other hand, \\y\\ < \\P\\ \\y + z\\ and ||z|| < \\IX - P\\ \\y + z\\, so
max(||y||,||.||)<(l + ||P||)||r(y,z )||.
(ii): If Y, Z are complete, then so is (Y 0 Z)^.
5.13 Let Y, Z be closed subspaces of a Banach space X. Show that X =
Y 0 Z (topological sum) if and only if X* =7i0Zi (topological sum).
Hint: If X = Y 0 Z, the statement follows from Exercise 5.9. Assume
X* = Yx 0 Z1. Then /(*) = 0 for all / E X* and x G Y n Z; hence
Y H Z = {0}. Using the separation theorem and Yx fl ZL — {0}, we find
that Y + Z is dense in X. We claim Y + Z \s closed. Let Q be the linear
projection of Y 0 Z (algebraic sum) onto Y. We claim that it is bounded.
Let P be the bounded projection of YL 0 ZL onto Y1. Fix y eY, z e Z.
Then, for x* Gl*, write z* = y* + ^* and
^(y)-y*(y)-y*^ + ^)<lb1llb + ^ll<l|P||lk1llb + ^l|)
so ||y|| < ||P|| ||y + *||. Thus ||Q|| < ||P||. By Exercise 5.12, Y®Z is closed.
5.14 Find two closed subspaces Y, Z of a Banach space X so that Y + Z
is not closed in X.
Hint: Consider Y = span{e2z_i}, Z = span{e22_i + 2~te2i} m X = l\.
Clearly, both are closed subspaces, but Y + Z is not closed. Indeed, zjsi —
N N N
J2 2~*e2; EY + Z since zN = ?(e2i-i + 2~ie2i) - J2 e2;-i- But zN -*
i—l » = 1 t = l
oo
z = X^ 2~2e2i, which cannot be in Y -f Z" because A,1,...) ??i.
i=l
5.15 Let Y, Z be closed subspaces of a Banach space X such that Y DZ —
{0}. Denote d = distEy, Sz)- Show that the subspace Y -f Z is closed in
X if and only if d > 0.
Hint: Assume that Y + Z is closed in X. Then Y is complemented in the
Banach space Y + Z, so there is a bounded linear projection ofY + Z onto
Z. For every y e SY, * G Sz, we have ||P||||y- z|| > ||P(y- z)\\ = ||y|| = 1,
and hence dist(?y, Sz) > H^ll-
Assume that Y-\-Z is not closed in X. By the first part of this proof, there
is no C > 0 such that C\\y — z\\ > \\y\\ for every y E Y, z ? Z. By scaling,
we have the existence of yn G^ and zn G ^, n E N such that ||yn|| = 1 and
\\yn+zn\\ < \ for all n. Hence |l-||zn||| = |||yn|H|-*n||| < ||yn + *n|| < ?¦
Therefore, by the triangle inequality, ||yn + H^nll^!! —> 0, so d = 0.

150 5. Structure of Banach Spaces
5.16 Let Y and Z be closed subspaces of a Banach space X such that
Y 0 Z = {0}. Define a norm |. || on Y 0 Z by |||y + z|| = ||y|| + \\z\\.
(i) Show that ||| • ||| is a complete norm on Y 0 Z'.
(ii) Show that the following are equivalent:
A) I • I is equivalent to the original norm on Y 0 Z.
B) Y + Z is closed.
C) y is complemented in Y + Z; that is, Y 0 Z is a topological sum.
D) There is k > 0 such that \\y\\ < k\\y + z\\ for every y E Y.
Hint: (i): {yn + zn} is Cauchy (resp. convergent) if and only if both {yn}
and {zn} are Cauchy (resp. convergent).
(ii): If I • I is an equivalent norm, then (Y 0 Z, || • ||) is complete, and
hence Y + Z is closed. But then Y 0 Z is an algebraic sum of two closed
subspaces of a Banach space; hence it is a topological sum. In particular,
Y is complemented by projection P such that \\y\\ < \\P\\ \\y + z\\. Finally,
the inequality implies \\z\\ < A + k)\\y + z||, so ||y + z|| < Bk + l)\\y + z||,
and clearly ||y + z|| < j||t/ + z|.
5.17 Show that there are two closed subspaces Mi and Mi of a Hilbert
space such that M\ C\ Mi — {0} and Mi + M2 is not closed, and thus there
is no projection on Mi that has kernel M2.
Hint: Define T E B{?2) by T(z) = B-*a?i) for x = (a:,-). Show that T maps
^2 onto a dense subset of ?2 but does not map it onto ?2.
Denote by G the graph of Tmt2®h, and let M = ^2 0 {0} C ?2 0 ?2-
Show that GflM = {0}, G+M is dense in^20^23 but that G+M ^ 4©^2.
5.18 Let X be a Banach space. Assume that Y, Z are u>*-closed subspaces
of X* such that Z©7 = X* (algebraic sum). Show that the corresponding
(algebraic) projection Q of X* onto Y is «;*-«;*-continuous.
There is a Banach space X such that an isomorphic copy Y of a Hilbert
space is a complemented subspace of X*; however, there is no w*-w*-
continuous projection of X* onto Y (see [JoLl]).
Hint: Note that Y± ®Z± —X (algebraic sum). Let P be the corresponding
(algebraic) projection of X onto Z±. Then P*(y + 2r)(yj_ -f ^j_) = Q(y +
^)(y± +^±) = 2/B:-l)-
5.19 Let X be a Banach space, and let Y be a closed separating subspace of
X\ For x e X, define ||x||y = sup{|y*(ar)|; y* E By} (c/., Exercise 4.15).
Show that:
(i) Y is norming if and only if X + Y1 is closed in X** (that is, it is a
topological sum).
(ii) Y is 1-norming if and only if X-\- YL is closed in X** and the projection
P of X + Yx onto X has ||P|| = 1.
Hint: (i): Assume Y is c-norming. Then, for x E X, yG^1, we have
Nl < c|M|y = c||z + y||Y < c||a; 4- y\\,

5. Structure of Banach Spaces 151
and the claim follows by Exercise 5.16. If the sum I0 71 is topological,
then X is isomorphic to (X + YL)/Y'L. Thus, for some c > 0 we have
?||*|| < p|| = dist(a?,Yx) = ||x||y.
(ii): If Y is 1-norming, by (i) we have \\x\\ < \\x + y\\. Now assume there is
a norm-one projection P of Z — X+Y1 onto X. Consider P: Zj Ker(P) —>
X. We have P(Bz/Ker(P)) — P(&z) — Bx and P is one-to-one, and hence
it is an isometry of Z/Kev(P) = Z/YL onto X. We get ||g|| = ||?|| =
dist(*,y-L) = \\X\\Y.
5.20 Let ?** G X** \ X. Show that Ker(z**) is 1-norming if and only if
||s** + ar|| > II^H for all x G X.
Hint: If Ker(***) is 1-norming, then the inequality follows by the previous
exercise. On the other hand, having the inequality, we obtain a norm-one
projection of X + Ker(***I = X + span{***} onto X (note that span{***}
is u>*-closed, so Ker(***I = (spanjz**}.!.I = span{z**}).
5.21 Assume that the codimension of X in X** is finite. Show that then
every t/;*-dense closed subspace in X* is norming.
Hint: Since X is finite-codimensional in X + 71, X + YL is closed in X**.
Use the preceding exercises.
5.22 Let y be a closed subspace of a Banach space, and recall that Y*
is isometric to X* jYL. Let q be the canonical quotient mapping X* —¦>
'X*/Y±. Show that if Z is a closed norming subspace of y* = X* /YL,
then W = q~x(Z) is a norming subspace of X*.
Hint: Assume without loss of generality that Z is a 1-norming subspace
of y* (consider || • || on Y and extend it to an equivalent norm in X,
Exercise 2.16). Take x G Sx. If dist(z,Y) < \, take y G Y such that
Ik-2/|| < i> so \\y\\ > f • Find V* ? Bz such that y*(y) > f. Thus, there is
u>* G §Pjy such that q(w*) = ?*; that is, iu*(i/) > |. Then x* = §w* G J9py
and x*(x) > |y*(y) — \ — \. Now assume that dist(#,Y) > |. Then
Ikll > | (norm in X/Y), so there is a:* G y1 such that #*(#) > |. Since
<?(?*) — 0 G Z, we get that x* G Pw, so in any case, for every x G Sx we
get sup{z*(x); a?* G P^} > |.
5.23 Let y be a closed hyperplane of a Banach space X. Show that for
any e > 0 there is a projection P of X onto Y with ||P|| < 2 + e.
In particular, X is isomorphic to Y 0 R.
Hint: Write X — Y © Z (algebraic sum), where Z = span(e). Using / G
(X/Y)* ^ YL such that /(e) = 1, find / G A + ?)?** such that /(e) = 1
and f(y) = 0 for y G Y. Let P(ar) = a: - /(a?)e.
The last part follows because Y is complemented.

152 5. Structure of Banach Spaces
5.24 Let y be a closed subspace of codimension n in a Banach space X.
Show that for every e > 0 there is a projection P of X onto Y such that
||P|| < n -f 1 + s. In particular, Y is complemented.
Hint: Let {/,-;#}? C (Yx x (Y1)*) be an Auerbach basis. Identify (Y1)*
with (X/Y)*\ Take a?t- G T{ with ||^|| < 1+e/n. Let P(x) = z-?/,(a:)a:2-.
5.25 Let Z be a, hyperplane of a Banach space X. Show that if there exists
a bounded linear operator T from some Banach space Y onto Z', then Z is
closed in X.
Hint: By taking quotients, assume that T\Y —> Z is one-to-one. Consider
the map 7\ from Y 0 R onto X defined by Ti(t/, r) = T(y) + r^o, where zq
is in the algebraic complement of Z in X. Then T\ G S(X 0 R, X) and is
onto, and hence it is an isomorphism, so T(Y) must be closed in X.
5.26 Let Y be a finite-codimensional subspace of a Banach space X. Show
that if there is a bounded linear operator T from some Banach space Z
onto Y, then Y is closed.
Hint: Induction on the previous exercise.
5.27 Let Y be a subspace of a Banach space X, and let xq ?Y. Show that
if Y is closed, then span(Y U {xq}) is closed in X.
This gives by induction that if Y is a closed subspace of X and F is a
finite-dimensional subspace of X, then span(Y U F) is closed in X.
Hint: Let / G X* be such that /(z) = 0 for all y G Y and /(z0) = 1.
Let yn + Anz0 ^ x e X, yn E Y. Then An = /(yn + Xnx0) -> /(ar).
Therefore, yn —> # — f(x)xo and a? — f(x)xo G Y since Y is closed. Then
x = x — f(x)xo + f(x)xo G span(Y U {a?o})-
5.28 Let X, Y, Z be Banach spaces, let T be a bounded linear operator
from X into Y such that T(X) is closed in Y, and let S be a finite-rank
operator from X into Z (that is, dim(S(X)) is finite). Define U: X —> Y0Z
by J7(a?) = (T(ar), 5(x)). Show that G(X) is closed in Y 0 Z.
Hint: If y*Gr,2*GZ*, and x G X, we have
tr(yV*)(aO = (y*>z*)(^(x))=(y*,0(r(^)^(x))
= ' y*(T(x))+,*E(ar))-(T*(y*) + 5*(.*))W,
so U*(y*,z*) — T*(y*) + S*(z*). Since S* is a finite-rank operator and
T* has a closed range by Exercise 2.39, we get that U* has a closed range
T*(Y*)-hS*(Z*) by Exercise 5.27. By Exercise 2.39, we obtain that U has
a closed range as well.
5.29 Show that the intersection of a finite-codimensional subspace and an
infinite-dimensional subspace in a Banach space is an infinite-dimensional
subspace.

5. Structure of Banach Spaces 153
Hint: The intersection of a subspace with a hyperplane is of codimension
at most one in the subspace. Then use induction.
5.30 Let X be a Banach space. Show that X 0 R is isomorphic to X if
and only if X is isomorphic to all its closed hyperplanes.
Hint: Assume that X is isomorphic to a closed hyperplane H. Then H 0R
is isomorphic to X 0 R and to X as well (Exercise 5.23).
If X is isomorphic to X 0 R by an isomorphism T: X 0 R —*• X, put
H — T(X 0 {0}) and observe that X is isomorphic to H, which is a closed
hyperplane in X. Then use Exercise 2.7.
The question whether every Banach space is isomorphic to its closed
hyperplanes was answered in the negative by Gowers ([Gowl]).
5.31 Let / E X*\{0}, where X is Co, c, or the space lp for 1 < p < oo. Show
that /~1@) is isomorphic to the original space. Equivalently, all closed
hyperplanes in cq or ?p are isomorphic to the whole space.
Hint: The closed hyperplane {@, x\, #2, • •.)] (%i) E X} in X is isomorphic
to X by the map (#i, #2? •••)*-* @, a?i, #2> • • •)] then use Exercise 2.7.
5.32 Show that the spaces Co and c are isomorphic. Recall that they are
not isometric (Exercise 3.75).
Hint: cq is a hyperplane in c given by the functional f(x) = lim(:rn).
Alternatively, consider T(x) — (lim(:rn), x\ — Yim{xn)) ?2 — lim(icn)J...).
5.33 Show that all closed hyperplanes of C[0,1] are isomorphic to C[0,1],
and C[0,1] 0 R is isomorphic to C[0,1].
Consider the subspace (C[0, l])o formed by all functions in C[0,1] that
vanish at 0. Use it to show directly that C[0,1]0R is isomorphic to C[0,1].
Hint: By the Sobczyk and Banach-Mazur theorems, there is an isometric
copy of Co complemented in C[0,1]; therefore, C[0,1] ~ cq 0 Z for some Z.
Show that the hyperplane {@, #i, #2, •. ¦)] (xi) E Co} 0 Z is isomorphic to
Co 0 Z. Then use Exercise 5.30.
For the second part: (C[0, l])o 0 R is isomorphic to C[0,1] by the map
(/, r) ^/ + r for/G(C[0,l])o,rGR.
5.34 Show that C[0,1] 0 C[0,1] is isomorphic to C[0,1].
Hint: Let Co be a subspace of C[0,1] formed by functions equal to 0 at 0
and let C\ be a subspace of C[0,1] formed by functions equal to 0 at 1.
By the previous exercise, Co, C\ are isomorphic to C[0,1]. Putting together
graphs of functions, we can see that C\ 0 Co is isomorphic to a subspace
of C[0,2] consisting of functions equal to 0 at 1. By scaling, this is in turn
isometric to a subspace C2 of C[0,1] consisting of functions that are 0 at
|. This is a hyperplane in C[0,1], and hence isomorphic to C[0,1] itself.
5.35 Show that C[0,1] 0 c0 is isomorphic to C[0,1].

154 5. Structure of Banach Spaces
Hint: Use previous exercises and the Pelczyriski decomposition method.
5.36 Show that Co is isometric to cq 0 cq with max-norm.
Hint: Map ((a?i, ar2, - - -)> (s/ii 2/2, - • •)) to (si>2/i>*2,2/2, • • •)•
5.37 Show that c 0 c is not isometric to c.
Hint: Assume that T is an isometry of c 0 c onto c. Consider x, y G (c 0 c)
defined by x = (A,1, • •-),(U, • • •)) and y = (A,1,...), (-1, -1, • • •))•
Then x,y are extreme points of Bc$c. Both hi = ^y2*-, h2 — ^^- have
norm one, and for each of them there exists an infinite-dimensional closed
subspace Y{ of c 0 c such that \\h{ + t/8-|| = 1 for every yi G Y% fl ?cec-
Isometry would have to carry these properties to c (in particular, it carries
extreme points to extreme points), but such behavior is impossible in c.
5.38 Show that C([0,1] U [2,3]) is isomorphic to C[0,1]. Note that [0,1] U
[2, 3] is not homeomorphic to [0,1] (connectedness).
Hint: Similar to the previous exercises.
5.39 Show that LP[Q, 1] 0^p is isomorphic to Lp.
Hint: Similar to the previous exercises.
5.40 Show that C^O, 1] is isomorphic to C[0,1].
Hint: Consider the mapping / h-» (/',/(())) and use that C[0,1] 0 R is
isomorphic to C[0,1].
5.41 Prove that a Banach space X is injective if and only if for every
superspace y of J, Banach space Z, and T G B{X) Z) there exists an
extension of T to Y; that is, some f G B(Y, X) such that f\x = T.
Hint: T = ToP. The other direction: use Z = X and T = IX.
5.42 Show that if Co C X and X is a separable Banach space, then there
is a projection P of X onto Co such that ||P|| < 2.
Hint: Simplify the proof of the Sobczyk theorem.
5.43 Recall that the distance d(x) of a point x = (x{) G ^oo to Co is equal to
limsup \x{\. Define a mapping <j> from l^ onto cq by ^(x)« = 0 if \x{\ < d(x)
and ^(a;),- — sign (a?*)(|a?*| — d(a?)) if \x{\ > d(x). Show that <f> is a Lipschitz
retraction of ^oo onto Co; that is, a Lipschitz map from t^ onto Co that is
the identity on Co.
Note that <f> cannot be linear because cq is not complemented in 4>o •
Hint: Direct calculation.
5.44 Write a formula for a projection of c onto cq.
Hint: P(x) = (xi - lim(xn), x2 - lim(arn),...).

5. Structure of Banach Spaces 155
5.45 Show that Co is not 1-complemented in c (i.e., there is no norm-one
projection of c onto Co).
Hint: Suppose that x = (x2) G c, lim(#2) ^ 0, is mapped onto 0 by a
i—*-oo
projection P. Let ||a?||oo = xno. Then ||a? - 2xnoeno\\ = xno and P(z -
2znoeno) = -2znoeno, so ||P|| > 2.
5.46 Let X be a Banach space. Show that there is a set T such that X is
isometric to a quotient of ^i(r). Show that there is a set T such that X is
isomorphic to a subspace of ^oo(r).
Hint: The proofs of Theorem 5.9 and Proposition 5.11.
5.47 We have conv(Ext(S^oo)) = Bi^. Show that ?& = ?\ contains a
u>*-compact convex subset C that is not equal to conv (Ext(C)).
Hint: Like any separable space, C[0,1] is a quotient of ?\ and we thus can
carry the unit ball of C[0,1]* into ?*. Then use Exercise 3.73.
5.48 Show that f% is isometric to a subspace of ?^ but not to any subspace
of c0.
Hint: Any separable Banach space is isometric to a subspace of ?&>
(Proposition 5.11). The space ?% has uncountably many extreme points, while
finite-dimensional subspaces of c$ have only finitely many of them. To see
the latter, use the uniform convergence to zero of elements of compact sets
in c0.
5.49 Let X be a Banach space. Show that if X is separable, then X* is
isometric to a subspace of ?<&.
Hint: The proof of Proposition 5.11.
5.50 Recall that ?\{c) is isometric to a subspace of C[0,1]* (Exercise 3.85).
Use it to show that ?\ (c) is isometric to a subspace of ?oq . Show that this
is not the case for co(c).
Note that ?^ contains an isometric copy of ?2(c) ([Ros2]).
Hint: Use Exercise 5.49. For the second part, there is no countable
separating family T in the dual of co(c). Indeed, if such an T exists, every
functional in T has a countable support, and hence there is r G c which is
not in (J supp(/). Clearly, f(er) = 0 for every / G T.
5.51 Suppose a Banach space X contains an isomorphic copy of ?\. Show
that then X* contains an isomorphic copy of ^iBN).
Hint: We have ^ as a quotient of X*. By the previous exercise, it has a
subspace isometric to ?i(c) — ^BN), so X* has a subspace Y such that
^iBN) is a quotient of it. By the lifting property, ^iBN) is a subspace of
y, and hence of X*.

156 5. Structure of Banach Spaces
5.52 Show that ^oo/co contains a subspace isometric to co(c).
Hint: Consider a system of characteristic functions of subsets A\ of N
with the property card(^A) — oo and card(^4\l C\ A\2) < oo whenever
Ai 7^ A2 (Lemma 5.16). Then the cosets containing xax form the standard
unit-vector basis of co(c) in the space Ioq /cq.
5.53 Show that t\ is of first Baire category in Co-
Hint: Btx is closed in c$\ if {ak} C Bix, ak = (af), and ak —>¦ a in c0, then
n
ak -^ a pointwise. Given n, we have ^2 |«f| < 1- Passing k —» 00, we get
i=i
n
^ |a,-| < 1, so a G JB/X-
*=i
Then, observe that 5^ does not contain any open ball of c$.
5.54 Show that I^tO, 1] is a set of the first Baire category in Li[0,1].
Hint: Bl2 is a weakly compact set in Li, so it is closed. Show by
constructing an appropriate function that Bl2 contains no interior point as a set in
Li.
5.55 (Lindenstrauss [Lin2]) Let X be a Banach space and assume that
(X 0 R) is isometric to a subspace of X. Show that X has a subspace
isometric to cq.
Hint: X contains a vector x\ E Sx and a subspace Y\ that is isometric to
X such that ||^i -f y\\ = max(l, ||2/||). Continuing by induction, we get for
every n G N a vector xn G /Sy^ and a subspace Yn-i of Yn isometric to X
such that \\xn + y|| = max(l, \\y\\) for every y G yn- It follows that for all
ll n II
real Ai,..., An we have J2 ^ixi\\ — max |Ai|. Hence, the closed subspace
of X spanned by {x{\ is isometric to Co.
5.56 (Lindenstrauss [Lin4]) Show that there is no separable reflexive
Banach space Xo such that every separable reflexive Banach space is isometric
to a subspace of X.
Szlenk proved that there is no separable reflexive Banach space X so
that every separable reflexive Banach space is isomorphic to a subspace of
X([Szl]).
Hint: Use the previous exercise for Xo and the fact that cq is not reflexive.
Note that this method works for many other classes of Banach spaces.
5.57 Let T be an abstract set. Show that ^i(F) has the Schur property.
Hint: Let xn ^ x in t\{T). Since xn G ^i(r), we have that Yl \xn{l)\ < 00.
Therefore, supp(xn) and supp(#) are countable. Let V = (Jsupp(xn) U
supp(#). Then V is countable, hence ^i(r') is isometric to l\. Since xn, x G
?i(r'), Theorem 5.19 implies that xn —* x.

5. Structure of Banach Spaces 157
5.58 Show that on Six the w*- and the norm topologies coincide.
Hint: By metrizability, it is enough to check sequences. Let #, xn G Stx and
* oo *
xn ^-+ x. Given e > 0, fix ko such that 53 \xi\ < e- Using xn ^-> #, find
i=k0 + l
ko ko oo
no such that 53 \x? ~xi\ < e f°r every n > n0. Since 53 |a?»|+ 53 lx*l ~ ^
2 = 1 z = l fco = l
Kq I fto &0 I
we also have 53 lxd > 1 — ?• Moreover, for n > no, 53 I^Pl ~ 53 \xi\\ ^
8 = 1 U' = l 2 = 1 '
fco ko ko oo
? |*?-x,-| < e and hence ? |x?| > ? k<l~e > i*- Since ? |x?| = 1
2 = 1 2 = 1 2 = 1 2 = 1
oo fc0
for every n, we also get ? l*?l = 1 ~ 53 \x?\ < 1 ~ i1 ~ 2*0 = 2e-
2 = fc0 + l *=1
OO OO
Therefore, for n > n0i ? |x? - s,-| < 53 (|x?| + |^-|) < 36: and
i=A;o+l i=k0 + l
oo &o oo
thus ? |x?-*,-|= ?|*?-a*| + ? |^-Xi|<4e.
2 = 1 2 = 1 2* = A;o + l
5.59 Let C be a weakly compact set in ^i(r). Show that C is compact.
Hint: By the Eberlein-Smulian theorem, C is weakly sequentially compact.
If {xn} is a sequence in C, there is a subsequence {xnk} that is weakly
convergent, and hence—by the Schur property—norm convergent.
5.60 Let X be an infinite-dimensional closed subspace of l\. Show that X*
is nonseparable.
Hint: 0 G Sx • If X* were separable, then (Bx,w) would be metrizable.
Thus there would be {xn} C Sx such that xn —» 0 in X, and therefore
xn ^ 0 in t\. By Schur's theorem, xn —> 0, a contradiction.
5.61 Let /, /i, /2,... G ?i[0,1]. Show that if fn —> / almost everywhere
and H/nld - H/lli, then /„ - / in Li[0,1] (Vitali).
Hint: \fn\ + \f\ - \fn - f\ -» 2|/| almost everywhere. Thus, by Fatou's
lemma,
2 / |/(*)| A < liminf / |/n(i)l + 1/@1 - l(/» - W)\di
Jo Jo
= 2 l |/(t)| (ft-limsup/ \(fn-f)(t)\dt.
Jo Jo
Hence limsup^1 |(/n - f)(t)\ dt = 0.
5.62 Let X be Co or ?p for p G [1, oo] \ {2}. Let T be an isometry of X onto
X. Show that there exists a sequence of signs a = ±1 and a permutation
7r of natural numbers such that T((xj)) = (e»a?7r(i)).

158 5. Structure of Banach Spaces
Hint: ?p case: Note that if vectors x,y satisfy \\x -f y\\p = \\x — y\\p —
\\X\\P + lbllP> then they nave disjoint supports. This is easy to see for p — 1
and requires elementary calculations for p > 1. Note also that this fails for
p — 2, since then any two orthogonal vectors satisfy the condition.
Using this, we see that T(e2) and T(tj) have disjoint supports for i ^ j.
Since T is onto, we readily obtain that supports of T(e2) must be singletons
and span{T(e2-)} — ?p.
If X — Co, we consider T**, which is an isometry of i^ onto l^, so we
use the previous result and T = T**|„.
5.63 Let X,Y be Banach spaces and let b be a map from 107 into R.
We say that 6 is a bilinear form on X 0 Y if b(x, yo) is linear in x for every
yo E Y and 6(zo, y) is linear in y for every #o E X.
Consider such a bilinear form b on X 0 Y. Show that the following are
equivalent:
(i) b is continuous at the origin @,0) of X 0 Y.
(ii) 6 is uniformly continuous on bounded sets of X 0 Y.
(iii) There is # > 0 such that \b(x, y)\ < K\\x\\ \\y\\ for all x E X and t/EY.
(iv) 6 is separately continuous onl0 7; that is, 6(z, yo) is continuous in
x for every yo E Y, and 6(zo, y) is continuous in y for every xo ? X.
Hint: (i) => (iii): Assume that there is 8 > 0 such that |6(#,y)| < 1
whenever x E <$i?x and y E <5By. By a homogeneity argument, we get that
\KX> v)\ ^ ^~2|kll llyll f°r everv ? E X and y EY.
(iii) :=> (ii): If ?, #' E #x and y, y' E ?y are such that \\x — x'\\ < 6 and
Hy-y'll < *, then
|6(^,y)~ 6(^,^I < |6(^-x/,y-y,) + 6(^,,y-y/) + 6(^-^/)y/)|
< *||*-x'||||y-j/|| + A%'||||y-y'||
+K\\j/\\ \\x - a?'|| < K82 + 2#«.
(ii) =$¦ (i) and (ii) => (iv) are trivial.
(iv) => (iii): Consider the family T — {b(xo, •); xq E Bx} of continuous
linear functionals on Y. For every yo E Y, we have sup |&(z,yo)| < oo
x?Bx
because this functional is continuous. Thus {&(zo, y); xq E #x} is bounded
for every y E Y, so by the uniform boundedness principle, sup{|6(a:, y)|; x* E
5j, y E i?y} < oo, which gives (iii).
5.64 Let p E [l,oo), x E lp. Let a > 0 and hj E <?p, j E N, be such that
\\hj\\ = a and hj ^ 0 in ^. Show that then \\x + hj\fp -> ||x||? + ap.
Hint: For z = (^) E <?p and n E N, denote in = (zi,..., znj 0,...) and
*n — @,... ,0,zn+i,zn_|_2, • • •)•
Given e > 0, from the uniform continuity of || • \\p on bounded sets there
is n such that |||s + /*j||? — \\xn + hj\\P\ < e for every j. Since hj -^ 0, there
is jo such that for every j > jo we have |||#n + ftj||? — ||^n + (^j)nllp| < e anc*

5. Structure of Banach Spaces 159
|H(^J)„H?-Il^-Il?| < e. Finally, note that ||*"H-(^J)„||? = ||S"||? + ||(Aj)„||?
for every j.
5.65 A function Q on a Banach space X is called a continuous quadratic
form if there is a continuous bilinear form 6 on X 0 X such that Q(x) =
b(x,x) for x G X and 6 is symmetric, that is, b(x,y) = b(y,x) for every
x}ye X.
Let p ? B, oo), and let Q be a continuous quadratic form on ?p. Show that
Q is w-sequentially continuous; that is, Q(xj) —* Q(x) whenever Xj -^ x in
Hint: Fixing ^E^ write Q(x + A) = Q(ar) + Q(h) + A(x} A) for he X.
Since A(x, A) is a continuous linear functional and thus weakly continuous,
we have that A(x,hj) —> 0 for every a?. We must prove that Q(hj) —> 0
whenever Ay —> 0. Assume that this is not true and for some e > 0 and
||Ay|| = 1, j G N, we have Ay A 0 and Q(Ay) > e for all j. Put ni = 1 and
^m = xi = Ai. Find n2 such that |A(a:ni, An2)| < § and ||arni + An2||? <
IknJIp + 2 (use the previous exercise). Put X2 = xni + An2, and find n^
such that \A(x2, An3)| < | and ||a?2 + ^n3||? < ||s2||? + 2, etc.
Then we have ||a?y||? < 2j for every j and
Q(xy) = Q(zy-i) + A(a:y_i, Ani) + Q(Ani)
> Q(xi_1) + e-|>g(a:i_1)+|.
Since Q(#i) = Q{h{) > e, we get by induction that Q(xj) > j\ for every
j. Then gM > i|Bi)/P = E^)^-2/* -> 00, which contradicts the
continuity of Q using the preceding exercises.
Similarly, all polynomials of degree smaller than p on ?p and all
polynomials on cq are weakly sequentially continuous ([BoFr]).

6
Schauder Bases
Definition 6.1
Let X be an infinite-dimensional normed linear space. A sequence {ei}^l1
in X is called a Schauder basis of X if for every x G X there is a unique se-
oo
quence of scalars (az)^l1; called the coordinates ofx, such that x — J2 aiei-
1 = 1
Clearly, {e2} is a linearly independent set in X.
If a Banach space X has a Schauder basis, then it is separable because
finite rational combinations form a countable dense set.
If X is n-dimensional, the notion of Schauder basis coincides with the
algebraic basis. We will write {ez}, span{ez}, etc., to cover both the finite-
and infinite-dimensional cases.
The condition in the definition can be weakened. If every vector x E X
n
has a unique decomposition J2aiei such that ]T) aiei —» ? as n —» oo, it
already implies that {e2} is a Schauder basis (see, e.g., [Sinl]).
If {e{} is a Schauder basis of a normed space X, then the canonical
(oo \ n
^2 cnei) = J2 aiei-
Lemma 6.2
Let {e{} be a Schauder basis of a normed space X. The canonical projections
Pn satisfy:
(i) dim(Pn(X)) = n;
yil) rnrrn = rmrn — rmin(m,n)/
(Hi) Pn(x) —> x in X for every x G X.

162 6. Schauder Bases
Conversely, if bounded linear projections {Pn}<^=1 in a normed space X
satisfy (i)-(iii), then Pn are canonical projections associated with some
Schauder basis of X.
Proof: The set {ez} is a linearly independent set in X. Thus (i) follows,
and (ii) is obvious. The property (iii) follows directly from the definition of
the Schauder basis.
Conversely, if projections Pn satisfy (i)-(iii), put Po = 0 and choose
eiGPi(X)nKer(Pi_i).Then
n
x = lim (Pn(x)) = lira (/>„(*) - P0(x)) = lim VY^x) - Pi-i(x))
%-\
oo oo
= 2(P<(x)-Pi_1(x))=^a,-e,-
i-\ i = l
for some scalars a2- because dim(Pn(X)/Pn-i(X)) = 1. The uniqueness of
oo
cti for x E X follows from the fact that \ix=Yl A e», then by the continuity
2 = 1
n
of Pn we get Pn(x) — J2 Piei> an(l hence #ej = P»(x) - P«-i(a?) = at-ei.
«=i
Thus, {e2} is a Schauder basis of X and Pn are projections associated with
{e.-}-
D
Fact 6.3
Let {ei} be a Schauder basis of a normed linear space X with canonical
projections Pn. 7/supn ||Pn|| < oo {we say that Pn are uniformly bounded),
then {e{} is also a Schauder basis of the completion X of X.
Proof: We will show that the extensions Pn of Pn on X satisfy (i)-(iii) of
Lemma 6.2. Since Pn(X) is finite-dimensional, it is closed in X and thus
Pn(X) — Pn(X), so (i) follows, (ii) is extended from Pn to Pn by the
continuity of Pn. Since Pn(x) —*• x for all x in a dense set X and Pn are
uniformly bounded, we have Pn{x) —> x in X, so (iii) is also true. Since
en e Pn(X) n Ker(Pn_i), we get
en E Pn (X)nKer(Pn_!) for every n.
Therefore, Pn are canonical projections associated with the Schauder basis
{efiofX.
D
Lemma 6.4
Let {e{} be a Schauder basis of a Banach space (X, || • ||). Define ||| • ||| on X
ii n ii oo
by |x|| = supn J2 aiei\\ for x = J2 aiei- Then:
(z) I • | is a norm on X, {ei} is a Schauder basis of (X, \\\ • |); and the
canonical projections Pn are uniformly bounded by 1 in ||| • |||;

6. Schauder Bases 163
(ii) I • I is an equivalent norm on X.
Proof: (i): The triangle inequality and homogeneity of ||| • ||| are simple to
check. Since for every x ? X we have ||a:|| = lim
n—>oo
? a*e*
i = l
by Lemma 6.2,
we obtain that |z| > \\x\\ for every x G X. This in particular means that
I • I is a norm on X.
To show that {ez} is a Schauder basis of (X, ||| • |), we use Lemma 6.2.
The properties (i) and (ii) are straightforward. To check (iii), we note that
for x G X we have
I* - Pm(x)\l = sup \\Pn(x) - PnPm(x)\\ = sup ||Pn(s) - Pm(ar)|| - 0
n n>m
as m —> oo. Finally, for m G N we estimate
|||Pm|| = sup lPm(x)i= sup sup||PnPm(;r)||=sup sup ||PnPm(ar)||
= sup{sup{||P„Pm(x)||; a: with sup ||Pi(a:)|| < l}}
< 1.
(ii): We will show that X is complete in the norm ||| • |||; that is, that
X C X, where X is the completion of X in ||| • |||. By (i), we already know
that {ez} is a Schauder basis of X. Given x G X) there is a unique sequence
of scalars &i such that x — Ylaiei-> wnere the convergence is in the norm
I • |||. Since ||| • ||| > || • || on X, we get that J2aiei ^s Cauchy in || • || and thus
convergent to some element x' G X in the norm || • ||. As shown in part
(i), J2aiei tnen converges to x' in the norm ||| • |||. Thus x — xl G X. This
means that X is complete in ||| • |||.
The formal identity map Ix' (X} \\\ • |||) —» (X, || ¦ ||) is a linear bijection of
a Banach space (X, ||| • |||) onto a Banach space (X, || • ||), which is continuous
since ||| • ||| > || • ||. By the Banach open mapping principle, it follows that
I^1 is continuous, which means that | ¦ ||| is an equivalent norm on X.
?
Theorem 6.5 (Banach)
Let {e2} be a Schauder basis of a Banach space X. The canonical projections
Pn associated with {ez} are uniformly bounded.
The value bc{e2-} = supn ||Pn|| is called the basis constant of {ez}.
Proof: Define ||| • ||| as in Lemma 6.4. Then |Pn||| < 1 for every n, and since
I • I is an equivalent norm, the result follows.
?
Considering the vectors en, we see that ||Pn|| > 1, in particular bc{e2} >
1. A Schauder basis {ez} is called normalized if ||ez|| = 1 for every n. It
is called monotone if bc{e2} = 1; that is, its associated projections satisfy
||Pn|| = 1 for every n.

164 6. Schauder Bases
Let {ei} be a Schauder basis of a Banach space X. For j G N and
CXI
x = J2 aieii denote fj(x) = a,. Then \\Pj(x) - Pj-i(x)\\ = ||/j(a?)e;-|| =
2 = 1
|/j(x)| • ||ej|| and thus
117,11= sup |/j(a:)| = Heyll sup H/^)^ < 211^11~x sup ||Pn||.
Therefore, fj G ^* for every j. The functional {/,-} are called the asso-
ciated biorthogonal junctionals (or coordinate junctionals) to {ez}, and we
oo
have x = ^ fi{x)ei for every rGl
i=l
We will denote the biorthogonal functional fi by e* and say that {e2; e*}
is a Schauder basis of a Banach space X. It is a biorthogonal system, and
we have just proved that ||e;|| ||e*|| < 2bc{ez}. Note that {e*} is separating
forX.
Examples
(i) Any linear basis of a finite-dimensional Banach space X is a Schauder
basis of X. In particular, any Auerbach basis is a Schauder basis,
(ii) Any orthogonal basis of a Hilbert space if is a Schauder basis of H.
(iii) If X — Co or X — ?p for p G [l,oo), then the sequence {e2} of the
i
standard unit vectors e2- = @,..., 0,1, 0,...) is a Schauder basis of X.
All these statements are easy to verify,
(iv) Let {tj}fL1 be a sequence of distinct points in [0,1] such that ti — 0,
t2 — 1, and {tj} = [0,1]. Define projections Pn from C[0,1] into C[0,1] by
Pi(f) = f@) and Pn(f) to be the piecewise linear function with nodes at
tj, j — 1,..., n and such that Pn(f)(tj) ~ f(tj) for j = 1,..., n.
By Lemma 6.2 and the uniform continuity of continuous functions on
[0,1], it follows that the projections Pn determine a monotone Schauder
basis of C[0,1]. This basis is called the Faber-Schauder basis of C[0,1].
(v) Let the functions hi be defined on [0,1] as follows: ho(x) = 1 for x G
[0,1]; fti(ar) = 1 for x G [0, \) and hx(x) = -1 for x G [\, l]; h2(x) = 1
for x G [0,|), A2(x) - -1 for x G [?,§], A2(*) = 0 for x G (|,l];
A3(x) = 1 for x G [|,|), A3(s) = -1 for x G (f,l], and ft3(ar) = 0
elsewhere, etc. Fix p G [l,oo). The set {hi} is linearly independent. Since
H = span{/i;} contains the characteristic functions of dyadic intervals, we
get?Lp = ?,[0,1].
For x = J2 aihi ? -H") define Pn(^) = Y2 ai^i (we can always assume
2=0 2=0
that m > n by adding 0/ij to the sum). These projections satisfy (i)-(iii) of
Lemma 6.2, so by Fact 6.3, to show that {hi} is a Schauder basis of Lp[Q, 1]
we must only prove that Pn are uniformly bounded on (H, || • || ).

6. Schauder Bases 165
n n+1
Assume that f = J2 ai^i an(^ 9 — S ai^i f°r some real numbers a;.
i=l 2=1
Then / and g differ only on some dyadic interval i", where / has a constant
value, say 6, and g has value b + fln+i on the first half of 7 and the value
6 — an+i on the second half of I. Since for every p G [l,oo) we have \b\p —
11F + an+1) + |F - an-i)|p < ||6 + an+i|p + ||6 - an+1\P by convexity of
|x|p, we get ||/|| < \\g\\ in Lp and it follows that the projections Pn have
norm at most 1. Thus {hi} is a monotone Schauder basis of Lp[0,1]. By
inspection, we see that {hi} is an orthonormal basis of ^[O,1].
(vi) On the space c, define projections Pn for x = (xi) G c by Pi(x) =
(a?i,xi,...), P2(x) = («i,»2,a?2,..-)» •••> ^0*0 = (xi,x2,...,z„,a:„,...).
Then, for # G c and n G N,
||z-Pn(z)|| = ||@,... ,0,sn+i - a?n, ^n+2 - xn,.. .)|| = max|sj - xn\ -> 0
J>n
as n —» oo. By Lemma 6.2, Pn generate a Schauder basis x\ = A,1,...),
n
*2 = @,1,1,...), ..., a?n = @,...,0,1,1,...). Given z = (zuz2:...) G c
and (ai,a2}.. .,an) such that (zi,z2, • - •, ^n, zn,...) = c*izi H hVn,
we calculate that z\ — ol\, z2 — cxi + a2, ..., zn — c*i -\ \-ctn. Therefore,
Pn{z) = »i^i + h anxn and ||Pn(z)|| = max|zj| — max{|ai|; |c*i+
j<n
a2\,..., |c*i + h an|}. The basis {xi} is called the summing basis of c.
Let {ei} be a Schauder basis of X. Considering ez- G X** we see that
{e*; ei} is a biorthogonal system in X*.
Fact 6.6
Ze/ {e2;e?} 6e a Schauder basis of a Banach space X with the canonical
projections Pn.
(i) For n G N, Pn*(/) = ? /(e^e* = ? e,-(/)e? /or every f ? X*.
2 = 1 2=1
(«) PnU) ^ / «« ** for every / G X*.
(m) {e*;e2} is a Schauder basis o/spanje*} with the canonical projections
P*. In particular, P*(f) —> / for every f G span{e*}.
oo
Proof: (i): For n G N, / G X*, and z = ? e*(z)e,-, we have
«=i
Pttf){*) = f(Pn(x)) = /(f>*(z)e,) = E/(ei)e;(«).
» = 1 i=l
(ii): By the continuity of f € X*,
n n
lira P'(/)(x) = lira ?>?(a:)/(e.-) = /( lira ?>*(z)et) = /(*)•
n—+ 00 n—»-oo *—» \n—>oo ^—» /
2 = 1 2=1

166 6. Schauder Bases
(iii): We easily check that P*P^ = P?in(ri|m). If / G span{e*}, then
P*(f) = / for n large enough and thus lim||P^(/) — /|| — 0. Since
||Pn|| = ||P^||, {Pn} are uniformly bounded and we can apply Lemma 6.2
and Fact 6.3.
?
Definition 6.7
Let {e2;e*} be a Schauder basis of a Banach space X. It is called shrinking
oo
if spanje*} = X*. It is called boundedly complete if J2 aiei converges
i = l
whenever the scalars ai are such that supn
J2 aiei
t=i
< oo.
Every normalized shrinking basis {e*} has the property that e» —> 0
because lim (et(e,-)) = 0 for every k G N. However, there exists a Banach
space with a normalized non-shrinking Schauder basis {e»} such that e8- —> 0
inX ([PeSz]).
Fact 6.8
Let {e2-;e*} fee a Schauder basis of a Banach space X with the canonical
projections Pn. The following are equivalent:
(i) {e2;e*} is shrinking.
(ii) {e*} is a Schauder basis of X*.
Proof: (i) => (ii): Fact 6.6.
(ii) =$> (i); If projections {P,*} generate a Schauder basis of X*, then
Pn(f) -* / for every / and thus X* = spanje*}.
(i) <=> (iii): Note that if P is a bounded linear projection of a Banach
space X onto P(X), then sup (/) = sup(P*(/)) = ||P*(/)|| and
P(BX) Bx
BP{x) C P{BX) C \\P\\BxnP(X) C \\P\\BP(X).
Thus we have, for every / G X* and n,
ll/|(/x_Pn)(X)ll = SUP{/(*); ^^%X"Pn)(X)}
< sup{/(x); xG(/x-Pn)(^x)}
< sup{/(x); xG(||Pn|| + l)p(/x_Pn)m}.
Hence \\f\{Ix_Pn){x)\\ < \\f - P*(/)|| < (||Pn|| + l)ll/|(/x_Pn)Wll- Thus,
{ei} is shrinking if and only if l|/L/x_P vxJI "^ ° for ever^ f ^ X* ¦
D
The standard unit-vector basis of Co and ?p, p G (l,oo), is shrinking,
whereas the standard unit-vector basis of l\ is not. The standard unit-

6. Schauder Bases 167
vector basis of ?p is boundedly complete for p G [l,oo), whereas the
n
standard unit-vector basis of cq is not, as the vectors A,..., 1, 0,...) show.
Proposition 6.9
Let {e2;e*} be a Schauder basis of a Banach space X. If {&i} is shrinking,
then the mapping T(x**) = {x**(ei)) ?5 an isomorphism of X** onto the
II n li
space of all sequences (a2) such that |(ai)| = suPn 1C aiei\\ < °°-
Moreover, if {e8-} 25 monotone, then T is an isometry.
Proof: It is routine to check that ||| • ||| defines a norm on the vector space of
(a,i) such that ||(az)||| < oo. Denote K = bc{et-}, and let Pn be the canonical
projections associated with {e,-}. For x G X, x* G X*, and x** G X**, we
have Pn*(z*) = ?>*(e,-)e? and Pn**(***)(**) = ? z**(e*K(e;), so we
«=i *=i
can write P,J*(x**) = ? ar**(ej)e,-. Thus
*=i
n
in***)! = sup||^^*«)ej| = sup ||/T (OH < # |IOI
2 = 1
and T is a bounded linear operator with ||T|| < K.
Now consider (a2)^:1 such that |||(at-)| < oo. Since X* is separable ({e2}
is shrinking) and < ]P aze2- > is bounded in X**, there exists a iu*-cluster
S=i J
point a:** of < J2 aiei >, which satisfies x**(e*) — a,. Moreover, ||e**|| <
limsup||f>ej < |(a,-)l. Thus, T(z**) = (a,-) and ||T(z**)|| > ||x**||,
Wi-i II
which completes the proof.
?
Theorem 6.10
Let {e2;e*} be a Schauder basis of a Banach space X. //{e2} is boundedly
complete, then X is isomorphic to (span{e*}) .
Proof: Let Pn be the canonical projections associated with {e2}. Denote
Z = span{e*} and define J:X —> Z* by J(x):z ^ z(a:); then J is a
bounded linear operator. We will show that J is an isomorphism of X onto
Z\
Let x G X. Then, for every z E Z, we have |«/(^)(z)| = \z(x)\ < \\z\\ \\x\\,
so ||«7(#)|| < ||a?||. On the other hand, for n G N, find #* G 5x* such that
x*(Pn(x)) = \\Pn(x)\\. Since P*(X*) = span{e?}?=i, we have P*(x*) €
Z, and ||Pn*(a;*)|| < K = bcfej. By definition, J (Pn(x)) (P*(x*)) =
(P*(x*))(Pn{x)) = x*{P*{x)) = x*(Pn(x)) = \\Pn(x)\\ and therefore

168 6. Schauder Bases
\\J{Pn(x))\\z- > J(Pn(x))(^fa) = p^)l[II^W|| > ?||Pn(*)||. By
the continuity of J, we have jf\\x\\ < \\J(x)\\z* < ||e|| for every x E X.
We will now show that J maps X onto Z*. To this end, observe first
that {e*, J(e{)} is a Schauder basis of Z, and let Pn denote its canonical
projections. Then P*(z*) ^ z* in Z* and supn ||P^|| = supn ||Pn|| < K <
oo; hence, for every z* E Z* and n E N, we have
|U(X;^(e?)e,-)| =|E**(eWte)|L.= lfoV)ll <*• 11**11-
ii \ / iiz ii iiz
i=l
Thus we have
E **(<*)*
»=i
»=i
<A'-
(f>*(e*h)
< A^2 • ||z*|L Since
the basis {e;} is boundedly complete, the series Yl z*(ei)ei *s convergent
2 = 1
in X to some xGX. Because J is a continuous map,
n n
J(x)= lim jQ>*(e*)e8) = lim ?>* (e*)J(et) = 1™ (?„*(**))
n—>-oo \*—' / n—»-oo ^—' n—»-oo v '
i=l
?=1
w*
in the norm topology of Z*. But i^(z*) -» z* in Z*, so 2:* = J(#) and J
is an onto map.
?
Theorem 6.11 (James; see, e.g., [LiT2])
Let X be a Banach space with a Schauder basis {ez}. X is reflexive if and
only if {ez} is both shrinking and boundedly complete.
Proof: Let X be reflexive. By Fact 6.6, for every / E X*, we have
Pn(f) ^ / and thus pn(f) -^ / in X* since X is reflexive. Therefore
X* — span™{e*} = span{e*} by Mazur's theorem, and {e2} is a shrinking
basis of X.
By Proposition 6.9, X** is isomorphic to the space Y = < (a2); |(aj)| —
II n II 1
max ^2 CLieAl < 00 >, Under this correspondence, X C X** corresponds to
n ||i=1 || J
Y\ — < (a;); Yl aiei converges >. Since X is reflexive, we get Y — Y\ and
{ez} is boundedly complete.
Conversely, if {e2} is shrinking and boundedly complete, then we have
the above identification and Y\ — Y; thus X — X**, and X is reflexive.
?
Definition 6.12
A sequence {e;} in a Banach space X is called a basic sequence if {ei} is
a Schauder basis of span{e2}.

6. Schauder Bases 169
A basic sequence {e2} is called shrinking if it is a shrinking basis of
span{e2}.
Proposition 6.13 (Banach)
Let {ei} be a sequence in a Banach space X. {e2} is a basic sequence if and
only if there is K > 0 such that for all n < m and scalars ai,..., am we
have
| 5Za*e»" - ^Ea»"e*
2 = 1
2 = 1
Moreover} the smallest such K is equal to bc{e2}.
Proof: One implication is clear from
n m m m
^flt-e,- = pn(X^aiCi) < ll^nll 5^a»ct- < bc{ej} ^at-ef-
2 = 1 2 = 1 2 = 1 2 = 1
On the other hand, suppose that K satisfies
J2 afeJ < K\\ ^ a,-eJ for
2 = 1
2 = 1
all
ai and n <m. We define projections Pn on span{ez} by Pn f J^ a2e2-) =
\=i '
n
J2 &it{ for m > n and scalars a2-, and observe that Pn have norm at most
2 = 1
K. We check that Pn satisfy (i)-(iii) of Lemma 6.2 on span{e;}, so by
Fact 6.3, {e{} is a Schauder basis of span {ei} and bc{e2} < K.
D
Not every separable Banach space admits a Schauder basis (Enflo; see,
e.g., [LiT2]). But we have the following statement.
Theorem 6.14 (Mazur)
Every infinite-dimensional Banach space contains a basic sequence. If X*
is separable, then X contains a shrinking basic sequence.
In the proof, we use the following lemma.
Lemma 6.15
Let Y be a finite-dimensional subspace of an infinite-dimensional Banach
space X. For every e > 0, there is x E Sx such that \\y\\ < (l-f?)||t/-f Ax||
for every y EY and every scalar X.
Proof: Let e G @,1). Let {y,-}^ be an |-net in 5y. For i E {l,...,m},
choose y* E Sx* with y*{yi) — 1. Since X is infinite-dimensional, there is
x E Sx such that y*(x) = 0 for every i — 1,..., m. We claim that x has
the desired property. Indeed, let y E Sy. Choose i E {1,..., m} such that
\\yi — y\\ < f * ^et ^ ^e a scalar. Then
\\y+ Xx\\ > \\yi + AarH - | > y*(Vi + Xx) - § = 1 - | > I^.

170 6. Schauder Bases
Thus, given y G Y \ {0} and a scalar A, we have ||]|*]r + fl^ll > 1^7-
D
oo
Proof of Theorem 6.14: Given e > 0, find en > 0 with [] A + ?n) <
n = l
1 + ?. Let xi G 5j be an arbitrary element. Using Lemma 6.15, construct
inductively a sequence {xn}%L2 m Sx such that, for every n > 1,
\\y\\ < (l + en)||y + Aa?n+i|| for all y G spanfai,..., xn}.
By induction, for n < m and scalars ai,..., am, we have
n mm
5^a*x* - A + ^)---(l + ?m-i)|^a^z| < A + e) ^aiXi
2 = 1 Z = l t = l
By Proposition 6.13, {x2} is a basic sequence and bc{#2} < 1 + e.
oo
Note that ||Pn|| < J"] A + ?«) —> 1 as n —> oo. If X* is separable,
i=n
we may assume without loss of generality that the norm of X is Frechet
differentiate (Theorem 8.19). By Proposition 8.34, the basic sequence that
we have constructed is shrinking.
?
It is not known whether every separable Banach space X contains a
closed subspace Y such that both Y and X/Y have a Schauder basis. We
also mention another open problem: If X is nonseparable, is there a closed
subspace Y of X such that X/Y is separable and infinite-dimensional? We
note that if X is separable and nonreflexive, there is a nonreflexive closed
subspace Y of X such that Y has a Schauder basis (Pelczyriski; see, e.g.,
[Dis2]).
Definition 6.16
Let {e2} be a basic sequence in a Banach space X} and let {fi} be a basic
sequence in a Banach space Y. We say that {e2} is equivalent to {fi} if
for all sequences of scalars (az); Ylaiei converges if and only if^ctifi
converges.
Fact 6.17
Let {ez} be a basic sequence in a Banach space X, and let {fi} be a sequence
in a Banach space Y. The following are equivalent:
@ {fi} is a basic sequence equivalent to {e2}.
(ii) There is an isomorphism T of span{e2} onto span{/2} such that
Tfa) = fi for every i.
(Hi) There are Ci, C2 > 0 such that for all scalars a\)..., an we have
~ n n n
i=l i=l i~\

6. Schauder Bases 171
Proof: (i) => (ii): Define a map T from span{e2} into span{/;} by
(CO \ OO
^2 cneA = J2 difi. From the equivalence of {ez} and {a?;}, we have
2 = 1 ' 2 = 1
that T is well defined, one-to-one, and onto span{/2}.
CO
We will now show that T has a closed graph. Indeed, if xk = ^ a*et-
2 = 1
CO CO CO
converge to x = ^ a^e; and T(zfc) = ]? a^/f converge to J^ cz-/2-, then
2 = 1 2 = 1 2 = 1
by the continuity of the coordinate functionals we have a* —» a2-, and in
the same way af —> c2- for every z. Hence a2- = c,- for every i, and thus
T(xA:) —» J2aifi — ^X^)- By tne cl°sed graph theorem, T is continuous,
and by the open mapping theorem, T is continuous as well.
(ii) => (iii): This follows easily with C2 = \\T\\, Cx = ||XT-1||.
II n I
(iii) => (i): For n < m and scalars ai,...,am, we have ^ a,-/,-
''2 = 1 '
II m II
C1C2 bc{e2} J2 a*/t , so by Proposition 6.13, {/;} is a basic sequence. By
i'j=i 'I
(iii), we also have that J2 aifi is Cauchy if and only if ^ aiei is Cauchy.
D
The following result is sometimes called the small perturbation lemma.
Theorem 6.18 (Krein, Milman, Rutman [LiT2], Valdivia [Val4])
Let {ei} be a basic sequence in a Banach space X, and let {e?} be the
coefficient functionals of the basis {ez} of span{e2}. Assume that {f} is a
CO
sequence in X such that J2 \\ei — fi\\ \\ei\\ = C < 1. Then:
2 = 1
@ {fi} Z5 a basic sequence in X equivalent to {ei}.
(ii) If span{ez} is complemented in X, then so is span{/z}.
(iii) If {e(} is a Schauder basis of X, then so is {fi}. Moreover, let {/,*}
be the coefficient functionals of the basis {fi} of X. Then span{e*} =
sparf{/;}.
Proof: (i): Extend e* to functionals on X of the same norm. For x G X,
CO CO
we have ? ||e?(*)(e,- - /*)|| < ||*|| ? IKII II* ~ Ml = C\\x\\, so S(x) =
2=1 2=1
CO
J2 e*(x)(ei — f) defines a bounded linear operator from X into X with
JJ5|| < C < 1. Let T = Ix ~ S. We have ||z - T(x)\\ = \\S(x)\\ < C||x||,
and hence \\T(x)\\ > A - C)\\x\\. Since 1 - C > 0, by Exercise 1.27, T is
an isomorphism into; in particular, T(X) is closed.
We claim that T(X) = X. Assume T(X) ? X. Since C < 1, by
Lemma 1.23 there is x G Sx such that dist(a?,T(X)) > C, which
contradicts \\x — T(x)\\ < C; so T is an isomorphism of X onto X. Then,
using T(ei) = /;, we obtain that T maps span{e;} onto span{/;} and (i) is
proved.

172 6. Schauder Bases
(ii): It follows from the proof of (i) using T and Fact 5.4.
(iii): Let T be the isomorphism of X onto X from (i). Since {e2} is a
Schauder basis of X, we get X = T(X) — T(span{e2}) = span{/2}, so {/;}
is a Schauder basis of X.
k
Fix some i G N and denote x*k = ]? /?(ej)e| for i;EN. Then z? G
j=i
span{ey} and, by Fact 6.6, #? ^-> /?. For z G B*-, we tnen nave /2*(a?) =
oo
J2 fi(ej)ej(x)> so f°r & > « we estimate
oo oo
K/;-*;)(*)! = | J2 ft(ej)e*(x)\ = \ ]T fite-fiVji*)
oo
< WftW E lki-/il|||c;|HOasi->oo.
i=fc+i
Since the estimate is independent of x G Bx, we get ||/* — z?|| —> 0 as
& —* oo and thus /* G span{e*}. Therefore span{/*} C span{e*}.
Note that (T-'Yie*) = /', so ||tf || < IRT)*!! ||e*|| for i € N. Thus,
the series J2 WfiW ||ei — fiW converges and we can reverse the roles of et- and
fi in the last paragraph, obtaining spanje*} C span{/*}.
D
Bases in Classical Spaces
Definition 6.19
Let {ei} be a basic sequence in a Banach space X. A sequence of nonzero
Pi+i
vectors {uj} in X of the form Uj — ]T c^e; with scalars a2- and p\ <
i=Pj + l
P2 < ... is called a block basic sequence of {ei}.
Note that a block basic sequence of {e^} is a basic sequence with basis
constant not greater than bc{ez}. Indeed, for k < /, we have
Pj+i
J=l j = l i=pj + l j = l i=pj + l
I Pj+i I
< bc{e2} HP E ajOje,- =bc{ei} ^QfjWj
i=i*'=Pj+i
i=i
The following result is often used when investigating subspaces of a given
space.

6. Schauder Bases 173
Theorem 6.20 (Pelczyriski [Pel2])
Let X be a Banach space with a Schauder basis {e2}. IfY is an infinite-
dimensional closed subspace of X, then Y contains an infinite-dimensional
closed subspace Z with a Schauder basis that is equivalent to a block basic
sequence of{e{}.
Proof: Let K = bc{e;}. Given pEN, let Wp be the finite-codimensional
subspace of X defined by
oo
Wp = Ix G X\ x = 2^ aiei\ ~ spa5T{ct}i>p-
i=p+l
Then Wp H Y is infinite-dimensional, so there is y G Sy fl Wp. We will
inductively construct the two equivalent basic sequences.
oo
Choose an arbitrary y\ = J2 a\ei ? ^ w^n I bill — 1- Find pi G N
*=i
pi
such that for u\ — Yl a\ei G X we have \\y\ — u\\\ < -gg. Choose y2 =
z = l
OO p2
X] a2^e« ? Sy H W^j, and fix /?2 E N such that for u2 = J2 rf^ we
have 11?/2 — ^211 < 2 . Continue in this manner. Then {uj} is a block
Zi ' Zi J\
oo
basic sequence of {e2}. Since ]T] \\yj— Uj\\ < ~jt and \\u*A\ < 2bc{uj} < 2i\,
by Theorem 6.18, {yj} is a basic sequence in Y equivalent to {uj}, so the
subspace Z = span{yj} has the desired property.
?
This procedure of finding vectors with almost successive supports is
called the "sliding hump argument." An easy modification provides the
following corollary.
Corollary 6.21 (Bessaga-Pelczyriski selection principle; see, e.g., [LiT2])
Let X be a Banach space with a Schauder basis {ei}. If a sequence {xn}
satisfies inf||#n|| > 0 and xn —» 0, then some subsequence {xnk} of {xn}
is a basic sequence equivalent to a block basic sequence of {ei}.
We will now investigate the structure of subspaces of ?p spaces.
Proposition 6.22
Let X be Co or ?p with p G [l,oo). // {uj} is a normalized block basic
sequence of the standard unit-vector basis {e;}; then {ui} is equivalent to
{ei}, span{i/2} is isometric to X} and there is a projection of norm one of
X onto span{uz-}.

174 6. Schauder Bases
Pi+i # Pi+i
Proof: We present the proof for ?p. Let t/j = J2 ^iei with ^ |AZ-|P —
1 for j E N. Then
m m Pj+i ± m Pj+i i
|X>HI = (E E kiw)? = (Emp E mp)'
j = 1 j = lt=pj + l j = l i=pj + l
m ± m
= (ElflilP)' = |^CaHI'
i=i j=i
so {e{} and {t/j} are equivalent and T(J2ajuj) — J2ajej *s ^ne isometry.
To find a projection, for every JEN choose u^ E span{ej}^ip1+1 C ?* for
which H^ll = Uj(uj) = 1. Then u^Uk) — 0 for fc ^ j and the operator P
oo
from X into spanjtfj} defined by P(x) — ]T u]{x)uj ls a linear projection
j"=i
Pi+i
of X onto span{i«j}. For a? = X^a«e* ? ^> we nave I^M^ < S la*'lP
«=/>j+l
for every j since ||u!-|| = 1 and thus
OO Pj+1 CO
\\p(*w = E E K*wip-i^ip = EKwr
j = li=pj + l j-\
oo Pi+i
< E E wp = nip-
This shows that ||P|| = 1.
D
Combining Theorem 6.20 and Proposition 6.22, we have the following
theorem.
Theorem 6.23
Let X be cq or ?p with p E [l,oo). IfY is an infinite-dimensional closed
subspace of X, then Y contains a subspace Z that is isomorphic to X and
complemented in X.
The method of the proof of the following result is called the Pelczynski
decomposition method.
Theorem 6.24 (Pelczynski; see, e.g., [LiT2])
Let X be cq or?p, p E [l,oo). IfY is an infinite-dimensional complemented
subspace of X, then Y is isomorphic to X.
Proof: Let X\ be a complement of Y in X\ then X is isomorphic to Y®X\.
Denote by ~ the equivalence of being isomorphic. By Theorem 6.23, there
is an infinite-dimensional subspace Z of Y such that Z ~ X and Z is
complemented in X, hence also in Y. Let Y ~ Z{&Y\. We will use the fact

6. Schauder Bases 175
that both 101 and (J2®X)x are isomorphic to X (Exercise 6.15). Thus
x 0 y ~ x e (z 0 Yx) ~ (x e z) 0 y1 ~ (x 0 x) 0 yx
~ x 0 Yx - z 0 Yi - y.
On the other hand,
X0Y - (iei®-)xey-((y®ii)©(y®ii) + -)x®y
- (iieii®---)i®(y©y®---)iey
- (Ii0ii0---)i0(y©y©---)x
- ((Y 0 Xi) + (Y 0 Xi) + • • -)x ~ (X 0 X 0 • - •)* ~ X.
Thus X - Y.
D
We note that for p E (l,oo), p ^ 2, there are closed subspaces of ^
that are isomorphic to ?p and not complemented in ?p. On the other hand,
if a subspace of ?p is isometric to ?p, then it is necessarily complemented
([Pel2]). Also, if p E [1, oo) \ {2}, then there are infinite-dimensional closed
subspaces of ?p that are not isomorphic to ?p. A reference for results in this
area is, for example, [PeBe].
When comparing the structure of Banach spaces, an important role is
played by certain classes of operators. Let X, Y be Banach spaces and
T E B(X,Y). We say that T is strictly singular if there is no infinite-
dimensional subspace Z of X such that TL is an isomorphism into Y. We
say that T is a compact operator if T(Bx) is compact in Y. Note that every
compact operator is strictly singular (Exercise 6.21).
Proposition 6.25 (Pitt; see, e.g., [LiT2])
Let 1 < p < r < oo. Every bounded linear operator T from ?r into ?p or
from Co into ?p is compact.
Proof: Assume that T:?r —> ?p is not compact. Since ?r is reflexive, there
is a sequence {xn} in ?r such that xn —+ 0 in ?r and ||T(#n)|| > e for
some e > 0 (Exercise 7.8). Then inf ||#n|| > 0, so using Corollary 6.21 and
Proposition 6.22 first in ?r and then in ?p, we find a subsequence {xnk} of
{xn} such that {xnk} is equivalent to the canonical basis of ?r and {T(xnk)}
is equivalent to the canonical basis of ?p.
If (ajt) E ?r \?p, then J2akxnk is convergent; hence T(^a^xn;c) =
Y^akT(xnk) E ^p. Thus J2 \ctk\p < oo, which is a contradiction.
The c0 case follows by Theorem 7.7.
?
We thus obtain that ?py?q are not isomorphic for p ^ q. Using
Theorems 6.23 and 6.25, we deduce that if p, q E [l,oo) and p / #, then
the spaces -?p and ?? do not have isomorphic infinite-dimensional closed
subspaces. Such spaces are called totally incomparable.

176 6. Schauder Bases
Theorem 6.26 (Dunford, Pettis, Pelczyriski; see, e.g., [DiUh])
Let X be a Banach space and T G B(co,X). If T is not compact, then
there is a subspace Z of c$ such that Z is isomorphic to cq and TL is an
isomorphism into X.
In particular, X contains a subspace isomorphic to cq. Before proving
this theorem, we fix some notation.
Let {xi} be a bounded sequence in a Banach space X. A sequence {yn}
is called an ^-average of {xi} if it is of the form yn = ^ a;?;, where
{An} is a sequence of pairwise disjoint finite subsets of N and ^2 |a2| — 1
i?An
for all n. Note that if {xi} is equivalent to the standard unit-vector basis
of ?i, then so is {yn}-
Lemma 6.27
Let {ez-,e*} be the canonical basis of cq. Let V be a bounded subset of ?\.
If V is not relatively compact, then there exist e > 0, sequences {xn} and
{yn} in V, a sequence of positive numbers {An}, and an ?i-average {zn} of
{el) such that for every n E N, \\xn— yn\\ > e and \\zn -An(xn — yn)\\ < \.
Note that then An < Jj for every n. Indeed, since ||zn|| = 1, the second
inequality implies ||An(sn - yn)\\ < f, while \\xn - yn\\ > e.
Proof: Since V is not relatively compact, there is e > 0 and an infinite
subset W of V such that \\x — y\\ > e whenever x,y G W, x ^ y. We
proceed by induction. Suppose that Xj, yj, Zj, A;- have been constructed
for j < n — 1. Let Aj = supply) and set Bn — [j Aj. Consider the map
j<n
T[Ylaiei) = J2 aiei acting from lx into Un = span{e;}2(E?n.
S = l ' i?Bn
Since Un is a finite-dimensional subspace of Co, we can cover Tn{V) with
finitely many balls of radius ^. Since W is an infinite set in V', we can
choose xn and yn in W such that T(xn) and T(yn) are in the same ball
and thus ||r(xn)-r(y„)||<|.
Find N > max(?n) such that \\(Itl -PN)(xn)\\ < ^, ||(I^ -PN)(yn)\\ <
Y~. Let Cn = {i; i > N} and An = N \ (Bn U Cn). Decompose xn —
xl + xl + xl and Vn= Vn + yl + Vn so tnat supp(a?),supp(^) C An,
supp(x2),supp(y2) c Bni and \\z*\\ < ^, ||j?|| < ^. Note that \\xn -
yn\\ = \\T(xn)-T(yn)\\<l^d
\\xl ~ yl\\ > \\Xn - Vn\\ - \\x\ - ^|| - ||a?|| - ||j?|| > fc/3.
Define An = ||^ _ ^y-i and ^ = A^2 _ ^2) Then supp(Zn) c An,
\\zn\\ = 1, and
\\zn - An(xn - yn)\\ = An||D - J/n) ~ (Xn ~ Vn)\\
< An(\\xn-yn\\ + \\xl\\ + \\yn\\3)

6. Schauder Bases 177
— 2e \6 ^ 12 ^ 12J ~ 2-
?
Proof of Theorem 6.26: (Bator [Bao]) If T is not compact, then also T*
is not a compact operator (Theorem 7.7). Because T*(Bx*) is not relatively
compact in ?i, from Lemma 6.27 we obtain e > 0, sequences {un} and {vn}
in Bx*> a sequence of positive numbers {An}, and an ^-average {zn} of
{e*} such that for every n G N we have ||T*(un) - T*(t;n)|| > e and
\\zn - An(T*(un) -T*(vn))\\ < ±.
For zn = ]T a2-e*, put Pn — {i G An\ ai > 0} and Nn = {i G An; a2- <
z€An
0}. Define xn G cq by #n = xpu ~ XNn for n G N. Then
(**,*„) _| Q ifiz^:
The space Z = {^tnxn] {tn} 6 Co} is isometric to Co because the #n's
have disjoint supports. Moreover, if {tn} G Co and ||{tfn}|| = \tj\ = 1 for
some j, then
\T(^2tnxn)\ = sup |**(T(^<„x„))|
[*,-(?*„*„)! - Ifo - A;(r*(^) -T*fa)))(E^n)[
2A;
so ||T(#)|| > §||#|| for x G Z and T is an isomorphism into.
?
We will now show that the structure of subspaces of Lp is different from
that of lp.
Theorem 6.28 (Khintchine; see, e.g., [LiT2])
Up ? A>°°); then Lp[0,1] contains a complemented subspace isomorphic
to ?2- ?i[0,1] contains a subspace isomorphic to ?2.
Note that Li[0,1] has no complemented subspace isomorphic to ?2
(Exercise 11.35).
In the proof of Theorem 6.28, we will use Rademacher's functions defined
by rn(t) = sign (sinBn7r/)) for t G [0,1] and n G N. Clearly, rn G Sz,p[o,i]
for every p > 1.
It is easy to observe that /0 ri(t)rk(t)dt = 8k\. Consequently, {rn} is
11 m 11
an orthonormal set in i^fO, 1]. In particular, ^ an^n = ||(an)||^25
n = l
L2

178 6. Schauder Bases
that is, {rn} considered in L2 is a basic sequence that is equivalent (with
constant 1) to the canonical basis of ?2- The behavior of {rn} in Lp spaces
is described in the following result.
Lemma 6.29 (Khintchine inequality)
Let rn be the Rademacher functions on [0,1]. For every p G [l,oo); there
exist positive constants Ap and Bp such that, for every ai,..., am;
m i_ p\ m « 1 m i
A,(Y,K\2y<( |e<w)| dty<Bp(j2K\2y.
n = l J° n = l n = l
By Ap and Bp we denote the best possible constants in this inequality.
They are called Khintchine's constants and their values are known. We
observed that A2 = B2 = 1. By the Holder inequality, it follows that if
p > r, then (? \f\p dt) P > ($ \f\r dtY. Consequently, Ar < Ap and
BT < Bp.
Proof: By the last remark, it is enough to show that there exist A\ > 0
and B2k < 00 for all k G N. We start with B2k-
fl\s^ \2k /Vv^ \2k
/ \Z^anrn(i)\ dt - I [2^anrn(t)\ dt
J0 'n = l ' J° n = l '
Jo
dt,
j
here the summation runs through all multiindices (ari,..., &j) with ]P a; =
2k and 1 < rci < ... < rij < m. By elementary combinatorics, .Aaif...>a. =
(a )?•¦•(a )\' Observe that fQ fj r%lk(t) dt = 1 if all oc{ are even, and it
ni<...<rifc
is equal to 0 otherwise. Using fa = a2/2, we therefore write
[l\ n i2Jk
rl, n
where the summation runs through subsets (/?i,..., f3j) of N such that
j
Y2 fa = k and 1 < n\ < ... < rij < m. Thus, we have
2 = 1
771 7
(Eki2) = !>* *e-<'
n=l
* """{xr^}E^ *.,<¦¦¦<'
wpu...apj

6. Schauder Bases 179
= min< > \)^anrn{t)\ at.
^A2CU...,2P3> Jo \^[ I
3
Since Yl A' — &> the minimum above is a minimum of a finite set of
2 = 1
positive numbers and so it is positive. Hence B2k exists finite.
m
To prove the existence of A\, put f(t) = Y2 anfn(t). By Holder's in-
n = l
equality used for p = | and g = 3, by the first part of the proof and our
observation before this lemma, we have
j\f{t)\2dt = j\f{t)\hfm"dt<{J\f{t)\dt)\j\f{t)\utf
a i i m i
1/@1 <ft)*a|(?l«»l2M
n = l
Therefore (/J |/(*)| dt) ' > 5^* (/„ |/(*)|2 <ft) s; that is,
/•l /*1 i m i
/ \f(t)\dt>B^( |/@I2*M = 54-2(X>„|2)T.
^° ^° n=l
Hence Ai > 5^.
D
Proof of Theorem 6.28: Define a map T from ?2 into Lp[0,1] by
oo
T((an)) = ^ anrn, where {rn} are the Rademacher functions. By
n = l
Lemma 6.29, it follows that T is an isomorphism from ?2 into Lp[0,1].
If p > 2, then ?p[0,1] is a subspace of L2[0) 1] (Exercise 1.8). Let P be the
restriction to Lp[0> 1] of the orthogonal projection of L2[0,1] onto span{rn}.
oo 1
We have P(f)(t) — ]T (JQ f(s)rn(s)ds)rn(t). This projection is bounded
in LP[0,1] since \\P(f%p < Bp\\P(f)\\L2 < Bp\\f\\L3 < Bp\\f\\Lr by the
Khintchine and Holder inequalities.
If p G A,2) and q > 2 satisfies \ -\- \ — 1, there is a subspace Y of
LjO, 1] isomorphic to ?2 complemented by a projection P. Then P*(Y*) is
a complemented subspace of LP[Q, 1] isomorphic to ^ (Exercise 5.10).
?
One of the consequences is that ?p is not isomorphic to Lp[0,1] for p ^
2, oo. Indeed, by duality, it is enough to show that ?p is not isomorphic
to Lp[0,1] for p < 2. Suppose that ?p is isomorphic to Lp[0,1] for some
p < 2. By Theorem 6.28, ?2 must be isomorphic to a subspace of fp, which

180 6. Schauder Bases
is a contradiction with Pitt's theorem. The situation is different if p = oo
because t^ is isomorphic to L^ (Exercise 6.18). Also, 1/2[0,1] is isometric
to ?2 by Theorem 1.38.
The spaces Lp and Lq are not isomorphic if p ^ q, py q G [l,oo). This
follows for instance using the notion of type and cotype. Ifl<jP<<7<2,
then ?p is not isomorphic to a subspace of Lq (this can be proved for instance
using the notion of type). However, Lp contains a subspace isometric to Lq.
We refer to [PeBe] for a survey in this area.
Unconditional Bases
Recall that a series ^ Xi in a Banach space X is unconditionally convergent
if J2eixi converges for all choices of signs a = ±1 (Exercises in Chapter 1).
Definition 6.30
A Schauder basis {e2} of a Banach space X is said to be unconditional if,
for every x E X, its expansion x = Ylaiei converges unconditionally.
A sequence {e;} in a Banach space X is called an unconditional basic
sequence if it is an unconditional basis of span{e2-}.
It is straightforward to check that the canonical basis {e;} of cq or ?pj
pG[l, oo), is an unconditional basis.
Also, as we saw in the notes following Theorem 1.38, every orthonormal
basis of a separable Hilbert space is unconditional. Note that every basis
equivalent to an unconditional basis is also unconditional.
Example
n
Let {ez} be the standard unit-vector basis of Cq. For n G N, set xn — ^ e;.
2 = 1
OO OO
We check that if x = J2 aiei> then x — Yl fin^n, where /?„ = an — an+i-
»=1 n=l
oo
It follows that {/?n}?=i forms a convergent series with ]T) j3n = ot\ and
n = l
I °° I
Halloo = s^Pfc ]C Pn • On the other hand, for every convergent series ]?/?*,
'n=fc '
oo oo / oo \
we have x = ]T Ckxk = ? ( ? Ph)ef- G c0.
fc=i t=i vjfc=; y
We just showed that (cq, || • H^) is isometric to the space of convergent
series {/3n} with the norm ||(/?n)|| = sup J ^ /?*
. The uniqueness of expan-
oo
sion x — J2Cnxn follows by induction, using the fact that ]P j3k — a2-, where
i—i
cti is the standard z-th coordinate of x (/?n is then necessarily an — an+i).
It follows that {xn} is a Schauder basis of Co, which is not unconditional.

6. Schauder Bases 181
Indeed, Y, n Xn ^ c°' wmle YnXn & c°- ^he basis {xn} is called the
summing basis of Cq.
Proposition 6.31
Lei {e2-} be a sequence in a Banach space X. The following are equivalent:
(i) {e{} is an unconditional basic sequence,
(n) There is a constant K such that for all scalars a\,..., am and signs
Si — ±1, we have
X^*a*e* - ^ Ea*e»
*=i
2 = 1
{iii) There is a constant L such thai for all scalars ai,
subset a of {1,..., m} we have
., am and every
i?a 2 = 1
We claim that the condition (iii) can be equivalently stated as
Y flte,*
<L
Y aiei
2 = 1
whenever a C N.
Indeed, let a- be any (even infinite) subset of N. Assuming (iii), we show
that Y aiei *s Cauchy: Given e > 0, there is no such that for m > n > tiq
i?cr
we have
Y aiei\\ < f;- Considering a' — a fi {n,...,m} and 6; = a2-
for n < i < m, bi = 0 otherwise, we use the condition (iii) to see that
< e. Thus Y aiei *s convergent. Passing to the limit, we
2*G<7
Y aieH
i?o\n<i<.m
get the claim.
However, in most applications, the finite-sum statement is easier to
handle because one does not need to discuss the convergence. A similar
observation can be made about the condition (ii).
Proof: (i) => (iii): Let Y = span{ez}. Given a C N, define an operator
PG from Y into Y by Pa(x) = Y a2e2- for z = Yaiei- The operator P^ is
i?cr
well defined since ^ aiti converges whenever Y aiei converges (use (i) and
Exercise 1.36). We now check that P0 has a closed graph. Indeed, let xk —> x
in Y for xk = Yaiei> a: = ^a*en and P<j{xk) = Y aiei -» 2/ = Y^i-
i i?a
From the continuity of the biorthogonal functionals in Schauder bases, we
have af —> ai for every z, and for the same reason, af —> 62- for every i.
Thus, 62- = a{ for every ? and hence Pa(x) = y, meaning that Pa has a
closed graph and is thus continuous.
Consider now the family of operators Pa, a running through all subsets
of N. We claim that for every fixed x = Yaiei ? X, the family {Pa(x)} is
bounded. Indeed, from the unconditionality we get that, given e > 0, there

182 6. Schauder Bases
is a finite set F C N such that J2 aiei < e whenever A O F = 0 (see the
exercises in Chapter 1). From this, the boundedness of {Pa(x)} for every
x E X follows. The Banach-Steinhaus uniform boundedness principle gives
that the operators Pa are uniformly bounded by some L.
(iii) => (ii): Given scalars ai,...,am and signs Si = ±1, we define
a = {i; Si = 1} and a' — {1,..., m) \ a. Then
I / J ?iQ>i?i
2 = 1
= Xlaiei ~ Yl aieil
m
— L^a*e* ~^~ zl-/ aiei\\ — ^ / ^fls-ej
z'E^
2Ga'
2 = 1
(h) => (iii): Given ai,..., am and a C {1,.
i E <r and ?t- = —1 for z E {1,..., m} \ a. Then
^ m
, m}, we define ?; = 1 if
i?o
2 = 1
1 m - m m
2 = 1
2 = 1
(ii) and (iii) => (i): Given n < m and scalars ai,..., am, we use (iii) with
<r = {l,...,n}to see that, by Proposition 6.13, {e2} is a basic sequence with
oo
bc{ez} < L. Now let ^ a2ez- be a convergent series. Given a — ±1, using
(ii) we show that J2 eiaiei
< 2K
Y2 aiei • Thus Yleiaiei *s Cauchy,
and hence convergent. This shows that Ylaiei converges unconditionally.
?
The best possible constant K from the condition (ii) in Proposition 6.31
is called the unconditional basis constant of {e2} and is denoted by ubc{e2}.
In the preceding proof we have shown that L < K and bc{e2} < ubc{e2}.
A natural question is whether every Banach space contains an
unconditional basic sequence (c/., Theorem 6.14). This long-standing problem was
answered in the negative by Gowers and Maurey ([GoMa]).
Theorem 6.32 (James; see, e.g., [LiT2])
Let X be a separable Banach space. If X has an unconditional Schauder
basis that is not boundedly complete, then X contains an isomorphic copy
of c0.
In the proof, we will use the following statement.
Lemma 6.33
Let {e;} be an unconditional basic sequence in a Banach space X. Then,

6. Schauder Bases 183
for all scalars (a2) such that Yaiei converges and all bounded sequences of
scalars {Xi}} we have
CO CO
2jAia,-eJ < ubc{ez-}(sup |A2|) ^a;ez- .
2 = 1
2 = 1
Y A;a2ez] = \\Y Aj-aiej
2 = 1 ' '12 = 1
and define a by e% = 1 if a2x*(e2) > 0 and a — — 1 if a2x*(e2) < 0. Then
2 = 1
2 = 1
5^At-at-et- < 5Z|Ai||a»x*(ei)| < ( sup |At-|) ]P?zazz*(ez)
1 " l<2<m
m
lv^ II
2 = 1
m
( sup |A2-|) ubc{ez} • y^a2eJ
l<2<m Hf-f '
< ( sup |A,-|)||x*
l<2*<m
2 = 1
D
Proof of Theorem 6.32: Let {e2} be an unconditional basis of X that is
ii n ii
not boundedly complete. Then there are scalars (az) such that Y flfej <
l'i=i 'I
CO
1 for every n and Y a2e2- does not converge. By the Cauchy criterion, there
2 = 1
are e > 0 and natural numbers pi < q\ < P2 < q2 '' • such that for Uj =
9i
Y a-j^j we have ||wj|| > e for every j, yet \\Y u\
j=pj
2 = 1
< K
Y aiei
2 = 1
<K,
where K — ubc{e2}.
By the previous lemma, for every sequence {\j}Jl1 of scalars, we have
772 771
I^A^I < ^(suplAyl) ^uj
i=l J i=l
< /^(sup^D-A^IKA^Hoo.
3
On the other hand, from the unconditionality of {e2} we have for each
|| 771 || c
i e {l,...,m}: \\Y xjuj\\ > jHlA*'u*H > j<\M\ that is> Tr||(Ai)||oo <
E XjUjl . Thus, {i/y} is equivalent to the canonical basis of cq.
j=i I'
D
Corollary 6.34
The space Li[0,1] does not have an unconditional Schauder basis.

184 6. Schauder Bases
Proof: By the classical Steinhaus theorem, Li[0,1] is weakly sequentially
complete (see, e.g., [Woj]). Let {e^} be the standard unit-vector basis of
n
Co. Then the sequence xn — Yl ei *s weakly Cauchy but not weakly con-
i=l
vergent, therefore Co is not isomorphic to a subspace of Zq[0,1]. If Z/i[0,1]
had an unconditional basis {#2}, {xi} would be boundedly complete by
Theorem 6.32 and thus Li[0,l] would be isomorphic to a dual space by
Theorem 6.10. But this is not possible: in a separable dual space, every
closed convex bounded set is the closed convex hull of its extreme points,
while the unit ball of Li[0,1] has no extreme points (Exercise 3.79).
?
Theorem 6.35 (James; see, e.g., [LiT2])
Let X be a separable Banach space. If X has an unconditional Schauder
basis that is not shrinking, then X contains an isomorphic copy of'l\.
In particular, a separable Banach space with an unconditional Schauder
basis contains an isomorphic copy of i\ if X* is nonseparable.
Proof: Let {ez} be the assumed basis; set K — bc{ez}. Since {e2} is not
shrinking, there is / G Sx* such that sup{/B:); z ? Bx nspan{e2}z>n} -f*
0. This means that there is e > 0 and a sequence n^ < n^+i such that
sup{/(z); z eBx nspan{enfc,enfc+i,...}} > e
for every k. Using the sliding hump technique, we can construct a
normalized block basic sequence Uj of {ea-} such that f(uj) > e/2 for every j.
Let m G N and ai,..., am be scalars. Assume ]P a; > Yl ~ai-
Then
z<m,ai>0
i < m, a i < 0
4Ea*w»
2 = 1
>
K
> 2
^2 aiUi\ -jc'f{ X^ aiUi)
z<m,ai>0 i<m,ai>0
ji E a^j E a' + ^ E -"»¦
z<m,ai>0 i<m,ai>0 2<m,a;<0
if E
i < m, a; > 0
>
^ <
e
X>i-
2 = 1
J2 —ai, we consider —a2- to obtain the same in-
i < 77i, a t < 0
equality. On the other hand, since {uj} is normalized, we have J2 aiui\\ <
'U'=i "
m it m it
J2 \ai\. Therefore, ^||(at)lk < Z) aiui < IIK)lk; that is, {uj} is
2=1^ ^ _ "i=l "
equivalent to the canonical basis of l\.
To see the second statement, note that if {e2; e*} is shrinking, then X* =
span{e2} and thus X* is separable.
D

6. Schauder Bases 185
Corollary 6.36 (James; see, e.g., [LiT2])
Let X be a Banach space with an unconditional basis. Then X is reflexive
if and only if X contains no isomorphic copy of ?\ or cq.
PROOF: It follows directly from Theorem 6.11 and Theorems 6.35 and 6.32.
?
Definition 6.37
The James space J consists of all sequences (a;) of real numbers such that
lim(af) = 0 and ||(a;)|| < oo, where the norm ||(a;)|| is defined by
||(at-)|| = sup f (ani - a„2J + (an2 - an3J + • • • + {ank_x - anJ2J 3.
ni<...<nk x '
It is standard to verify that (J, || • ||) is a Banach space. Moreover, the
n
sequence {e$} of the standard unit vectors en — @,..., 0,1, 0,...) is a
monotone Schauder basis of J.
Claim
{e,-} is a shrinking Schauder basis of J and J** = J © span{(l, 1,1,...)};
in particular, J is not reflexive.
Proof: Assume that the basis is not shrinking. As in the proof of
Proposition 6.35, we find x* G J*, e > 0, and a normalized block basic sequence
{uj} of {e{} such that x*(uj) > e for every j. Consider the vector u =
oo
J2 iwj • By Holder's inequality applied to the norm of J and by considering
7=1 J
the division points of u;'s, we find that ||u||2 < 2^ ^||^||2 < 2J^ -^ < oo.
On the other hand, x*(u) = ^2x*(uj)/j is not finite, a contradiction.
By Proposition 6.9, J** is identified with (/%) such that supj. Yl foei
k
i=l
<
oo via the map T(x**) = (a?**(cj)). From the definition of the norm II • II of
II k II
J and from the fact that sup J ^ A'^i < oo, it follows that lim (/%) = f3
exists. Thus (/?« —/?)» corresponds to an element of J since lim (pi—ft) = 0.
2—»-00
Therefore J** = J 0 span{(l, 1,...)}.
?
Since J** is separable, J does not contain an isomorphic copy of cq or
l\. Indeed, if J contained an isomorphic copy of Co, then J* would have a
quotient isomorphic to cj, and J** would have a subspace isomorphic to
cj* = ^oo. A similar argument shows that J does not contain an isomorphic
copy of t\.
By Corollary 6.36, J does not have an unconditional basis.
However, J contains an isomorphic copy of ?2] in particular, it contains an
unconditional basic sequence.

186 6. Schauder Bases
Bessaga and Pelczynski proved (see, e.g., [LiT2]) that if a Banach space
X has an unconditional basis and Y is a nonreflexive closed subspace of X,
then Y contains an isomorphic copy of Co or l\. Using this result and the
fact that J (like every separable Banach space) is isometric to a subspace
of C[0,1], we see that C[0,1] does not admit any unconditional basis.
It was a long-standing problem whether every Banach space contains
either a reflexive subspace or an isomorphic copy of cq or t\. This problem was
answered in the negative by Gowers. In fact, Gowers showed ([Gow2]) that
there is a separable Banach space X such that every infinite-dimensional
closed subspace of X has a nonseparable dual and yet X contains no
isomorphic copy of t\.
Theorem 6.38 (Bessaga, Pelczynski; see, e.g., [LiT2])
Let X be a Banach space with no subspace isomorphic to Cq. If for a
sequence {?„}??_! in X there exists M > 0 such that ]P snxn
n = l
< M for
every m G N and every choice of signs en = ±1, i = 1,..., m, then Ylxn
is unconditionally convergent.
The condition is often equivalently stated as J2 \x*(xn)\ < M for every
n = l
x*eSx..
Indeed, given x* G Sx*, set en — 1 if x*(xn) > 0 and en = —1 if
x*(xn) < 0. Then, for m G N,
mm km
]P |x*(a?n)| = Ylx*FnXn) ~ x*{%2€nXn) - lYlCnXn
n — 1 n — 1 n — 1 n = l
On the other hand, assume the condition with x*. Given en = ±1, m G
N, we find x* G Sx* such that
7 j ?nxn
n = l
( \
Vn = l y
and then
IX^^^I ^ ]Ci**(^)i - M-
n = l
n = l
Proof: We will show that J2xn is convergent. Since we can repeat the
proof for every E ?nxni ?n — il> we get that ^2,xn is unconditionally
convergent. Assume that ]T)zn does not converge. Then, by the Cauchy
criterion, there is e > 0 and integers pi < qi < pi < qi • • • such that
ik
E Xn
n=Pk
> e for every k. Put yk
Vk
E xn for every fc. Then ]? |/(t/fc)| <
oo
? |/(a?n)| < Afll/H for every / G XMn particular, /(</*) -> 0; that is,
Jb = l
t/jk —* 0. Moreover, e < \\yk\\ < 2M for all k; hence there is a subsequence
(denoted again {yk}) that is a basic sequence (Corollary 6.21). Consider

6. Schauder Bases 187
J2 dkVk. Then
m m
< ^ajb^fc =sup| ^ajb/(yfc) ; / G BX*J
Jfc = l Jk = l
m
< (max|aib|)sup{?; \f(yk)\; f € Bx*} < M^at)^
k=i
Hence, {yk} is equivalent to the canonical basis of Co, a contradiction.
bcjyjb}
D
Note that the canonical basis of cq satisfies the condition in Theorem 6.38
and is not unconditionally convergent.
Theorem 6.39 (Bessaga, Pelczynski; see, e.g., [LiT2])
Let X be a Banach space. If X* has a subspace isomorphic to cq, then
X has a complemented subspace isomorphic to l\. In particular, X* has a
subspace isomorphic to t^.
Proof: Let T be an isomorphism from cq into X*. Then T* maps X** onto
l\ (Exercise 2.39). Since Bx is w*-dense in Bx** by Goldstine's theorem,
using Corollary 2.25 we find K > 0 and xn G X for n G N such that
n-l
lkn|| < K, T*(xn)(en) - 1 and ? |T*(zn)(ez-)| < ?, where e,- are the
2 = 1
standard unit vectors in cq. By Theorem 6.38 and Proposition 6.22, it
follows that {T*(xn)} has a subsequence {T*(xnk)} that is equivalent to the
canonical basis of ?\ and whose span is complemented in l\ by a projection
P. Therefore, for some constant M > 0 and every choice of scalars {ak}^=1
satisfying ?^ \a,k\ < oo, we have
oo
YlakXnA
Jb = l
oo
<kJ2m
k=i
oo
Jb=l
oo
Hence T* is an isomorphism of Y = span{znfc} onto span{T*(#nfc)}. Thus,
Y is isomorphic to l\ and P = {T*)~1PT* js a projection of X onto Y.
Finally, P*(X*) is isomorphic to P(X)* =?*=?<» (Exercise 5.10).
D
Given a bounded sequence in a Banach space with a separable dual, we
can extract a weakly Cauchy subsequence. Rosenthal characterized spaces
sharing this property as spaces not containing an isomorphic copy of l\.
Precisely, he proved that if {xn} is a bounded sequence in a Banach space
X, then either {xn} has a weak Cauchy subsequence or contains a
subsequence that is equivalent to the standard unit-vector basis of ?\ (see, e.g.,

188 6. Schauder Bases
[LiT2]). There are separable spaces whose dual is nonseparable and yet
they do not contain an isomorphic copy of l\ (e.g., the space JT discussed
in the exercises).
Markushevich Bases
Definition 6.40
Let X be a Banach space. A biorthogonal system {xa\ fa}aer in X is called
a Markushevich basis of X if span{a?a}aer = X and {/a}aer separates
the points of X.
A Markushevich basis {xa]fa}aer is called shrinking if span{/a} — X*.
Clearly, every Schauder basis of a Banach space X is a Markushevich
basis of X. An example of a Markushevich basis that is not a a Schauder basis
is the sequence of trigonometric polynomials {e22?rnt; n = 0, ±1, ±2,...} in
the space C[0,1] of complex continuous functions on [0,1] whose values at
0 and 1 are equal, with the sup-norm.
Theorem 6.41 (Markushevich; see, e.g., [LiT2])
Let X be a separable Banach space. If {z{}{ C X satisfies span{2j}j = X
and {gi}i C X* separates points of X, then there is a Markushevich basis
{xi\ fi} of X such that span{x2-} = spanjzf} and span{/2} = span{<7;}.
Proof: Define x\ — z\ and f\ — gk1/9k1{zi)i where k\ ? N is such that
gk1(zi) ^ 0. Then find the smallest integer /12 such that gh2 ? span{/i}.
Define f2 - gh2 - 9h2{xi)fi- Find an index k2 such that f2{zk2) / 0,
and set x2 ~ (zk2 — fi(zk2)xi)/f2(zk2)- Let hz be the smallest integer
such that Zh3 ? spanjzi, x2}- Put x3 = zh3 - fi(zhz>)xi - f2(zhz)x2 and
h - (9k3 -9k3(xi)fi -9k3(x2)f2)/gk3(x3), where k3 is an index such that
gk3(xs) ^ 0. Continue by induction. At the step 2n, we construct /2n
first; at the step 2n + 1, we start by constructing a?2n+i- It follows that
span{^}" C spanl^-}^ and span{#2}" C spanj/;}^. Clearly, fi(xj) =
6{j, span{xi} C span{z2} and span{/2} C span{<7;}.
D
It is an open problem whether every separable Banach space X admits
a Markushevich basis {#;; fi}i^i with \\xi\\ — \\fi\\ = 1 for all i. It is known
that, given a separable space X and e > 1, a Markushevich basis of X
exists so that sup \\xi\\ ||/2|| < 1 + e (Ovsepian-Pelczyiiski; see, e.g., [LiT2]).
Theorem 6.42 (Gurarii, Kadec [GuKa])
Let Z be a closed subspace of a separable Banach space X. Any
Markushevich basis {xf, fi) of Z can be extended to a Markushevich basis of
X.
Precisely, there are Zj ? X, <pj ? X*, and extensions of fi to ipi ? X* such
that {{xi} U {zj}] {i/ji} U {^j}} is a Markushevich basis of X.

6. Schauder Bases 189
Proof: Extend all /; onto X and denote these extensions by fi. Let {yj, </>j}
be a Markushevich basis of X/Z (it is separable). For all j, choose yj E yj
and define <fj{x) — ^i(^) f°r x ? -^j no^e that ^;'(x0 — 0 f°r a^ z- We have
span{{#z} U {%}} — X and {/;} U {<?>;} is a family separating points of X.
3 ~ i
Put Zj - yj - Yl ^ijxi and fa ~ U - Yl ^ijVji where A2J- = /»•(%•) for
t=i i=i
i ^ j and A22- = \fi{yi)> Then {{#z}U{zj}; {^}U{<^j}} is a Markushevich
basis of X that extends {#;, /;}. Indeed, from the definition of Zj and ipi, it
is clear that spanja;;, Zj} = X, {^z} U {<?>j} is separating points of X, and
V>i extend fi onto X. It is routine to check that the system is biorthogonal.
?
Theorem 6.43 (Johnson, [Johl])
The space ioo does not admit a Markushevich basis.
We will need the following lemma.
Lemma 6.44
Let X be a Banach space. Every reflexive subspace of X* is w*-closed in
X\
Proof: Let Y be a reflexive subspace of X*. By the Banach-Dieudonne
theorem, it suffices to show that By is w*-closed in X*. Since By is
incompact in y, it is ^-compact as a subset of X*. Therefore, By is
incompact in X* and thus w*-closed in X*.
?
Proof of Theorem 6.43: Assume that {xa]fa}r is a Markushevich
basis of ^oq. Put Y = span{/a} r. We claim that Y is reflexive. It is
enough to prove that By is weakly sequentially compact.
Let {yn} be a sequence in By. Since every element of span{/a} has a
countable support over {xa} (as a limit in norm of a sequence of points in
span{/a} that have finite support), there is a countable subset TV of T such
that for a E T \ N and n E N we have yn(xa) = 0.
By the Cantor diagonal argument, let {ynk} be a subsequence of {yn}
such that lim (ynk(xa)) exists for every a E N. Since yn(xa) = 0 for every
k —>-oo
a E r \ N and {yn} is bounded, we have that lim (ynk(x)) exists and is
k —+oo
finite for every # E X. Denote this limit by y(x). Then y is a bounded
linear functional on l^ and ynk —»?/ in l^. By the Grothendieck property
of loo (see Exercises 6.30 and 6.31), this means that ynk —> x/ in ^ and
y E By since ?y is ^-closed in i^. Therefore, By is u>-compact in t^ and
thus also w-compact in Y. Consequently, Y is reflexive.

190 6. Schauder Bases
By Lemma 6.44, Y is w*-closed in l^, and since Y is separating for
-?oo, it is w*-dense in i^. Thus Y — t^. Consequently, t^ and thus i^ is
reflexive, a contradiction.
?
Theorem 6.45 (Plichko)
Let X be a Banach space. If X is separable, then X* is a complemented
subspace of a Banach space with Markushevich basis.
In particular, Iqq is a complemented subspace of a Banach space with
Markushevich basis.
Proof: Let I be a set with card(J) = dens(X*). Define U = fcco)
and Z = (X* ®[/)r Let {en; fn} be a Markushevich basis of X scaled so
that ||en|| < 1 for every n. Choose a dense set {yl}i?i in Bx*. For i ? I
and n G N, let
n
? = •-&>;)?•
For a fixed n G N, define hn — {{yl(en)}k} ¦• We have y\(en) — 0 for k > n
and |*4(en)| < 1 for fc < n, and hence
V~ ^oo(r)
Let uln be the standard unit vector in U and gln be the standard unit
vector in the space (X^i) = U*. For n G N and i G /, put xln —
(VniK) € z and e'n = (en,-/in) G Z*. We claim that {{zjJneN.iei U
{(/m30)}n,meN;{e^}meN U {@,^)}n6N(ie/} is a Markushevich basis of
Z.
It is easy to see that this is a biorthogonal system and that the functionals
separate points of Z (efn = en on I*).
It remains to show that the span of vectors {xln} fj}n,jeN)iei is dense in
Z. To see this, we note that for n G N and i ? I we have
n n
< + y{ = < + ? +?*/'>;)/; - < + ?</>;)/; € spanK,/,}^.
Also, for every i G /, we have
E
<+!/* ,11 1
n — l
- > 0 as k
Therefore y% G span{a^,/;}. Furthermore, for every n G N and 2 G /, we
n
have < = < - sA = < - </ + ? </(e;)/; ^ span{4, /j}-

6. Schauder Bases 191
Thus spa.n{xln,Xj}n = Z, and the proof is complete.
?
Exercises
6.1 Let {e7} be a Hamel (i.e., algebraic) basis of an infinite-dimensional
Banach space X. Show that some of the coordinate functionals associated
with this basis are not continuous.
Hint: Pick an infinite sequence {eni} in {e7}. Consider the vector x =
OO g
^2 2~z n* . Since {e7} is a Hamel basis of X, we have x — Y^xieh
*=1 llenill F
where F is a finite set. Let np be such that np (fc F, i.e., xUp = 0. For
every m, ^ 2~2 n* has the np-coordinate equal to 2~np/||enp||. If the np-
i=l \\eni\\
coordinate functional were continuous, we would have xUp — 2~nr/\\enp\\ ^
0, a contradiction.
6.2 Why do we not use (iii) in Lemma 6.2 and the Banach-Steinhaus
theorem to conclude that Pn are uniformly bounded?
Hint: Do we know that they are bounded operators?
6.3 Show that the canonical projections of a Schauder basis of a normed
space X need not be uniformly bounded if X is not a Banach space.
Hint: Consider the trigonometric polynomials in the space of continuous
functions on [0,27r].
6.4 Use the notion of basic sequence to prove that a Hamel basis of an
infinite-dimensional Banach space has cardinality at least continuum.
Hint: Take any basic sequence {xn} in X. Let {Na}a G T be a collection of
infinite subsets of N such that Na n Np is finite if a ^ /? and card(r) > c
(Lemma 5.16). Define ya — Y2 2-i?;. Then {ya} is a linearly independent
i?Na
set of cardinality at least c.
6.5 Let {e2} be a Schauder basis of a Banach space X. Prove that there is
an equivalent norm on X in which {e;} is monotone.
Hint: Put |z||| = supn ||Pn(a?)||. Then
|Pm(x)| = SUp ||PnPm(x)|| = SUp ||Pn(s)|| < SUP ||P„(x)||.
n n<m n
6.6 Let {e$} be a Schauder basis. For n < m G N, define Tn>m ( J2 aiei J =
m
Y2 a-i^i- We say that {e2} is bimonotone if ||Tn>m|| — 1 for all n,m.
i—n

192 6. Schauder Bases
(i) Show that ||r„,m|| < 2bc{e;}.
(ii) Show that there is an equivalent norm ||| • ||| on X such that {e;} is a
bimonotone basis of (X, ||| • |||).
Hint: (i): T„,m = Pm - Pn-i-
(ii): |z| = sup ||T„,m(z)||.
6.7 Let {e2} be a Schauder basis of a Banach space X. Show that {e^} is
monotone if and only if ||Pn+i(#)|| > ||Pn(x)|| for every x ? X and n ? N.
Hint: If ||Pn|| = 1 f°r all n> use Pn(%) — PnPn+i(ff)- If the condition holds,
prove first ||Pfc(a:)|| > ||Pn(z)|| for all k > n and then take the limit for
k —> oo.
6.8 Show that C[0,1] has a Schauder basis consisting of polynomials.
Hint: Consider any Schauder basis {e2} of C[0,1] and approximate {ei}
by polynomials using the Stone-Weierstrass theorem. Use the stability
theorem for Schauder bases.
6.9 Let X, Y be closed subspaces of a separable Banach space Z such that
XHY = {0}. Assume that the algebraic sum X-\-Y is not closed in Z. Show
that there exist a basic sequence {xi} C X and a basic sequence {yi} C Y
such that \\xi\\ = ||y2|| = 1 and \\xi — y2|| < 4~z for n ? N. In particular, X
and Y have infinite-dimensional closed subspaces that are isomorphic.
Thus, if X and Y are totally incomparable spaces (for example, ?p,lq for
p ^ </), then X + Y is closed in every overspace.
Hint: If X + Y is not closed, then (Exercise 5.15) there exists x ? X, y ? Y,
with 11a?11 = ||r/|| = 1 and \\x — y\\ arbitrarily small. Then use the proof of
Proposition 6.14 and Proposition 6.18.
6.10 A Banach space X is said to have the approximation property if for
every compact set K in X and every e > 0 there is a bounded linear
finite-rank operator T from X into X such that \\x — T(x)\\ < e for every
x ? K. Show that every Banach space X with Schauder basis has the
approximation property.
Note that a different definition of this property is given in Chapter 7
after Proposition 7.4. The proof that they are equivalent can be found, for
example, in [LiT3].
Hint: Let {Pn} be the canonical projections for some Schauder basis of X.
Then we have lim ||Pn(#) — x\\ = 0 for x ? X. The limit is uniform on
n—>-oo
compact sets since supn ||Pn|| < oo. See the proof of Proposition 7.4.
6.11 Let {e2;e*} be a Schauder basis of a Banach space X. Show that if
{ei} is shrinking, then {e*} is a boundedly complete basis of X*.

6. Schauder Bases 193
Hint: {e*} is a Schauder basis of X*. Consider a; such that sup ]T) a2e*
"i=l
<
oo. X is separable, so there is a sequence {n&} C N and x* ? X* such that
xnk — J2 aiei ~* x* (Exercise 3.111). Since {e*} is a basis of X*} we
2 = 1
oo
have x* = ? Ae*. Fix j ? N. Then lim(ar*fc(ei)) = x*(ej) = fa and
i-\ k
oo
z* (e;) = Oj for rik > j] hence Oj = fa. Thus ? at-e* = x*.
2 = 1
6.12 Let {e2;e*} be a Schauder basis of a Banach space X. Show that
span{e*} is a norming subspace of X*.
Hint: Let \\x\\ = sup||Pn(s)||. Then |Pn|| = 1 for every n, so |||Pn*||| = 1.
Given / ? Sx*, we have lflPn*(/)l < 1, P*(f) ? span{e*,..., e*n}t and
P*(f) —> /• Hence span{e*} fl J5^* = i?^*, where 5^* is the unit ball
in the dual norm to ||| • |||. Thus, span{e*} is 1-norming for (X, ||| • |||). Since
I • I is an equivalent norm on X, span{e*} is a norming set.
6.13 Let X = (X^^oo)o- Show that every infinite-dimensional closed
subspace of X contains an isomorphic copy of ?2.
Hint: Given a closed subspace Z of X, there is a subsequence in Z that is
equivalent to a block basic sequence of the canonical basis of X such that
the "nodes" are at the ends of the canonical blocks. This block basis is
equivalent to the standard unit-vector basis of ?2.
6.14 Show that (X^?p[0,1]), is isometric to LP[0,1].
Hint: Put An = [^j, ^] and identify / ? Lp[0y1] with the sequence {f\A }.
Use the isometry of Lp[0,1] and Lp(An).
6.15 Show that {J2^p)i ls isometric to ?p and (X^co)c is isometric to cq.
Hint: Direct computation.
6.16 Show that gC[0,1]) is isomorphic to C[0,1].
Hint: Use Pelczynski's decomposition method. Represent (X)C[0,1]) as
a complemented subspace of C[0,1] using an infinite number of nodes in
[0,1] and the closed subspace of C[0,1] of functions that vanish at these
nodes.
6.17 Let p ? A, 00). Show that Lp[0,1] 0 ?2 is isomorphic to Lp[0,1].
Hint: Use that ?2 is isomorphic to a complemented subspace of Lp[0,1].
6.18 Use Akilov's result that Loo[0,1] is complemented in every overspace
to show that i^ is isomorphic to Loo[0, 1] ([Pell]). Because ?\ is not
isomorphic to Li[0,1], this provides an example of two non-isomorphic spaces

194 6. Schauder Bases
whose duals are isomorphic. This also shows that no isomorphism of t^
and ?00[0,1] can be w*-w*-continuous (Exercises 2.41 and 3.20).
Hint: As the dual of a separable space, Loo is isomorphic to a subspace
of ^00 • ^00 is complemented in every overspace, so we use the Pelczynski
decomposition method.
6.19 Show that ?2 is isomorphic to a quotient of ?co. Compare this result
with Exercise 11.33.
Hint: The space ?2 is isomorphic to a subspace of ?i[0,1] by Theorem 6.28.
Therefore, ?\ — ?2 is isomorphic to a quotient of L\ — L^, and then use
the previous exercise.
6.20 Assume that T is a bounded linear operator from a Banach space
X into X such that T(X) is not closed. Show that there is no finite-
codimensional subspace Y of X such that T would be an isomorphism
from Y into X.
Hint: Assume that T is an isomorphism from a finite-codimensional Y
into X. We have X ~ Y ^ Z since Y is complemented. Then T(X) =
T(Y) + T(Z). If T is an isomorphism on Y, then T(Y) is closed. Since Z
is finite-dimensional, we have that T(Y) -f T(Z) is closed.
6.21 Let X, Y be Banach spaces. Show that every compact operator from
X into Y is strictly singular.
Give an example of a strictly singular operator that is not compact.
Hint: Let Z be a subspace of X such that T is an isomorphism of Z
onto T(Z). Then T(Bz) is a closed subset of T(Bx)\ hence it is
compact. Because T\„ is an isomorphism, Bz must be compact; hence Z is
finite-dimensional.
For the second question, consider the formal identity map from ?\ into
e2.
6.22 (Kato) Let X be an infinite-dimensional Banach space, and let T be
a bounded linear operator from X into X such that the restriction to every
infinite-dimensional closed subspace of X is not compact. Show that there
is a finite-codimensional subspace Z of X such that the restriction of T on
Z is an isomorphism.
Hint: Assume that the restriction of T to every finite-codimensional sub-
space is not an isomorphism. Then, for every 6 > 0 and {a?*}^ C X*,
there is x E Sx with ||T(a?)|| < 6 and x*(x) = 0 for i = 1,..., m. By
the proof of Mazur's theorem, there is a basic sequence {xi} C X with
bc{xi} < 2 so that \\T(xi)\\ < 8~f. Let Z = span{zf-}. Then T(BZ) is
contained in the set {Y^a>iT(xi)\ \ai\ < 2}, which is totally bounded due
to ||T(^)|| < 8_i (see Exercise 1.48).

6. Schauder Bases 195
6.23 Show that the sum of two strictly singular operators T, S from a
Banach space X into a Banach space Y is strictly singular.
Hint: By repeated use of the previous exercise, first for T and then for 5,
get an infinite-dimensional closed subspace Z of X on which both T and
S are compact. Then T + S is compact on Z and thus not an isomorphism
on Z.
6.24 Let T be a strictly singular operator from a Banach space X into a
Banach space Y. Is T* necessarily strictly singular?
Hint: No. Consider a bounded linear operator T from ?\ onto ?2
(Theorem 5.9). T is strictly singular because ?2 is not isomorphic to any subspace
of ?\ (Pitt's theorem). However, T* is an isomorphism into (Exercise 2.39)
and thus T* is not strictly singular.
6.25 Let T be uncountable and 1 < p < q < 00. Show that there is no
bounded linear one-to-one operator from ?q(T) into ?p(T). Similarly, there
is no bounded linear one-to-one operator from cq(T) into ^p(r), p G [1, 00).
Hint: Assume that such an operator T:?q(T) —> ?p(T) exists. Let e > 0 and
an uncountable set Ti C T be such that ||T(e7)|| > e for 7 G IV An infinite
sequence in {e7}7Gr! tends weakly to zero, and we use Pitt's theorem for
its closed span.
6.26 Show that every strictly singular operator T from ?p into ?p is
necessarily compact for p G [1, 00).
Hint: Examine the proof of Pitt's theorem.
6.27 Does there exist a bounded linear operator from C[0,1] onto ?2?
Note that there is no isomorphic copy of ?2 complemented in C[0,1] (this
follows using the Dunford-Pettis property; see Chapter 11).
Hint: Yes. ?\ is isomorphic to a subspace of Za[0,1], which is isomorphic to
a subspace of C[0,1]* by the Riesz representation theorem. By Lemma 6.44,
?*2 isw*-closedinC[0,l]*.
6.28 Does there exist a bounded linear operator from ?p onto tq, p ^ q G
[1,00)?
Hint: No. Pitt's theorem, duality.
6.29 Does there exist a bounded linear operator from ?p onto Co?
Hint: If and only if p = 1. For p — 1, see Theorem 5.9. For p > 1, use the
reflexivity of ?p.
6.30 A Banach space X is said to have the Grothendieck property if for all
{#*} C X*, z* ^ 0 implies z* ^ 0.
Show that if X has the Grothendieck property and T is a bounded linear
operator onto Y, then Y has the Grothendieck property.

196 6. Schauder Bases
Hint: Direct proof, use the fact that T* is w*-w*-continuous and T** is
onto (Exercise 2.39).
6.31 Does there exist a bounded linear operator from i^ onto Co? Does
there exist a bounded linear operator from t^ onto ?\1
Hint: No. It is known that l^ has the Grothendieck property ([Dis2]) and
Co does not.
Assume there was a bounded linear operator from l^ onto ?\. By
Theorem 5.9, Co = <?(^i), so we would get a bounded linear operator from l^
onto Co-
6.32 Show that a block basic sequence of an unconditional basis is
unconditional.
6.33 Show that every normalized unconditional basis of a Hilbert space is
equivalent to the canonical basis of ?2.
Hint: Let J^Xi be unconditionally convergent. For e > 0, there is no
11 m 11
such that J2 ?iXi\\ < e for every n,ra > no and a = ±1. The gener-
II m n2 m
alized parallelogram equality says ^ 1C ?ixi\\ — 2m~n+1 ^ \\xi\\2,
?r,=±l"n=:l " n — 1
m
so J2 \\xi\\2 < e2!. Thus, if {e2-} is a normalized unconditional basis of a
i—n
Hilbert space H, then ^a;e; converges only if Ylai converges.
On the other hand, if J^ a? < 00, then from unconditionally and the
parallelogram equality we get that J2aiei ls unconditionally convergent.
Note that an analogous result holds for Co and l\. However, in these cases
the proof is more difficult (see, e.g., [LiT2]).
6.34 Let X be a Banach space. Let {xn} be a bounded sequence in X such
that {xn} has no weakly Cauchy subsequence. Prove that X contains an
isomorphic copy of l\ (which follows from Rosenthal's theorem mentioned
earlier in the text) assuming that X has an unconditional basis.
Hint: X* must be nonseparable; otherwise we use the Cantor diagonal
procedure to produce a weak Cauchy subsequence. If X* is nonseparable,
use Theorem 6.36.
6.35 A Banach space X is called weakly sequentially complete if every
weakly Cauchy sequence is weakly convergent in X. An example of a weakly
sequentially complete space is any reflexive space (show this) and any
general Li(fi) space (this is a classical Steinhaus theorem). Show that if X
is weakly sequentially complete and nonreflexive, then X must contain an
isomorphic copy of l\.
Hint: The first problem: use the Eberlein-Smulian theorem. The second
problem: If X does not contain an isomorphic copy of ?1, every bounded

6. Schauder Bases 197
sequence {xn} in X has a subsequence {xnk} that is weakly Cauchy by
Rosenthal's theorem. Since X is weakly sequentially complete, {xnk} is
weakly convergent in X. Therefore, X is reflexive by the Eberlein-Smulian
theorem.
6.36 Let X be a Banach space (not necessarily separable). A family
{e7}7er in X is called an unconditional Schauder basis of X if for
every x E X there is a unique family of real numbers {a7}7er such that
x — ]Pa7e7, where the summation is meant in the sense that for every
e > 0 there is a finite set F C T such that
< e for ev-
~ E a7
-yeF'
ery F' D F. Note that for every x E X only countably many coordinates
a7 are nonzero. Indeed, given n E N, there is a finite set F C T such
that ]T a7e7 < - for every finite F1 disjoint from F. Applying this to
ll7€jp/ II
F' — {7} and assuming ||e7|| = 1, we get {7; |a7| > ^} C F.
Similarly as for the case of countable Schauder bases, it can be shown
that the coordinate functionals a* are in fact in X*.
Show that the closed linear span of the coordinate functionals of an
unconditional basis is a norming subspace in X*.
Hint: Use P*.
6.37 Let X be a Banach space. Show that if X* is separable, then X admits
a shrinking Markushevich basis.
Hint: The proof of Theorem 6.41.
6.38 Let X, Y be separable Banach spaces. Show that there is a bounded
linear operator mapping X onto a dense subset in Y.
Hint: Let {xi\fi} and {zi\Qi} be Markushevich bases of X and Y,
respectively, with {fi} and {zi} bounded. Put T(x) — ]T^2~2fi[x)Zi.
6.39 Let X be an infinite-dimensional separable Banach space. Show that
there is a biorthogonal system \x%\fi\ such that spanjz;} — X and {/;} is
not separating.
Note that this cannot be done in finite-dimensional spaces.
Hint: Let {yi} be a linearly independent sequence such that span{yz} = X.
Pick x E X\span{y2} (infinite-dimensional spaces cannot have a countable
Hamel basis; Exercise 1.64). Since {x, yi,...} is a linearly independent set,
using Hahn-Banach we find gi E Sx* such that gi(x) — 0, gi(yi) — 1, and
9i(Vj) — 0 for j = 1,..., i — 1. Then {gi} separates points of span{t/2}
and as in Theorem 6.41 we find a biorthogonal system {xi\ fi] so that
spanjxf} = span{2/f} and span{/;} = span{^}. Then spanjzf} = X and
{fi} does not separate points. Indeed, gi{x) — 0 for all i; hence also f(x) — 0
for all / E span{#;}.

198 6. Schauder Bases
6.40 Show that James's space J (Definition 6.37) contains an isomorphic
copy of ?2.
Hint: Let H denote the subspace of J consisting of vectors with even
coordinates equal to zero. Show that on H, James's norm is equivalent to the
?2 norm.
6.41 Let Ui = ei — e2-_|_i, where {e2} is the shrinking basis of J. Show that
{ui} is a boundedly complete basis of J, which is not shrinking, and
m-1 "j+i-1 2 i
]C&W^=SUH(^( ^T Zi) j ; 1 < ni < ... < nmj.
j = l i=zrij
6.42 Let {ui} be the boundedly complete basis of J from the previous
exercise. Let {v{} be the biorthogonal functionals to {«»}, which span the
predual space J* to J by Theorem 6.10. Define g E J* by g(^2?iUi) — ]T)&.
Show that J* = span({x;2} U {g}) — span(J* U {#}).
Hint: J* = span{?;2}. Clearly, g E J* \J*] then use dim(J*/J*) = 1.
6.43 Assume that we replace the exponent 2 in the definition of James's
space J by 1. Is the resulting space isomorphic to ?{l Can cq be obtained
in a similar way?
Hint: Yes. Yes.
6.44 Is the following an equivalent norm on James's space J?
m
Mlo= sup {?I***-* ~ *n*l2}-
ni<...<n2m^JT1 }
Hint: Yes.
6.45 Define a norm ||| • | on James's space J by
1x1 =SUp—=((?ni -Xn2J + ..- + (xnm_1 - XnmJ + (xnrn - XnJ2)* .
Show that I • I is an equivalent norm on J and, in this norm, J** is
isometric to J.
Hint: Consider the map U: J** —> J defined by
U(x") = (-A, x"{ei) - X, ***(e2) - A,...),
where A = lim(#**(e*)).
6.46 Find a Schauder basis {et-} of a Banach space X with separable dual
such that {e{} is not shrinking.
Hint: Consider the biorthogonal functionals {/;} to the standard unit-
vector basis {ei} in James's space J. Since the standard basis is shrinking,

6. Schauder Bases 199
{fi} is a basis of J*. However, {/;} cannot be shrinking because its
biorthogonal functionals are {e2} in J and span{e2} — J ^ J**.
6.47 Does there exist a Banach space X not isomorphic to a Hilbert space
and such that X* is isomorphic to XI
Hint: Consider {lp 0^J, where p G (l,oo) and - + - = 1. To obtain a
nonreflexive example, consider J 0 J*, where J is James's space.
6.48 Does there exist a Banach space X such that X(&X is not isomorphic
toX?
Hint: Consider James's space. What is the codimension of J0J in (J(BJ)**?
In the remaining exercises, we define and investigate the James tree space
JT. All notations and definitions made in one exercise carry to the following
ones. For more information, see [FeGa].
6.49 Let T - {(n, i); 0 < i < 2n - 1, n G N0} be equipped with a partial
ordering < determined by the relation (n, i) < (n + 1, j) iff j G {2i, 2i-f 1}.
Then (T, <) is a binary tree. By a segment we mean a subset S = {t G
T; (n,z) < t < (m,j)} for some (n,i),(m,j) G T. A maximal linearly
ordered subset of T is called a branch. Let T denote the set of all branches
in T. Show that card(F) = c, the cardinality of the continuum.
6.50 The James tree space JT consists of all real functions defined on T
with the norm
k 2 i_
3 = 1 (n,i)eSj
where the supremum is taken over all finite sets of pairwise disjoint
segments in T. Verify that JT is a Banach space with a boundedly complete
basis {e(nj2)} ordered lexicographically; that is,
@,0)<A,0)<A,1)<... < (n,z)<(n,z + l)<...<(n,2n>-l)
< (n + l,0)< ...
and the functions e(n2) are defined by e(nti)(mij) = 1 if (ft, i) — (m,j) and
zero otherwise (i.e., e(n>o(m>i) = S(™,j)(n,i)).
In particular, there exists a predual JT* to JT with a shrinking basis
{f(n,i)} with e(nJ)(/(mj)) = 6(m}j),(nti)- Thus, JT* can be represented as a
oo 2n-l
space of real functions F on T satisfying F = if*- lim ?^ ^ ^(^3 0/(n,*)-
n=0 z=0
Hint: By contradiction. Consider z G JT and ? > 0 such that for
every n G N there exists a finite set {S?}j=1 of pairwise disjoint segments
satisfying ( ]T ( ]?) ?(n,i)J J > ?. Then there exists a fast-growing
yj=i\n,i)es? J J

200 6. Schauder Bases
sequence {n/} of integers such that 5/ * C\ Sj 2 = 0 for l\ ^ h- Thus
Y^{ ]C x(n)z)J = oo, a contradiction.
' \n,i)esp J
6.51 For every branch 7 G T, let Y7 — {x G JT; supp(x) C 7}. Show that
Y1 is a subspace of JT isomorphic to James's space J.
Show that the linear operator P7: JT —> JT defined by
00 2n-l
n=o i=o te-y
is a norm-one projection of JT onto Y.
Hint: See Exercise 6.41.
6.52 For F G JT\ define 5(F) G *2(r) by S{F) = ( Km F(n,«))
n—>oo,(n,z)G7 ^
for 7 G T. Show that S: JT* —> ^(r) is a bounded linear mapping.
Hint: \\F\\ — sup{F(a;); ar G Bjt}. Using this, one can get ||5|| — 1. To
show that S is onto, given {ji}f=1 C T and scalars ai,..., an, find F G JT*
such that 5(F) = JT ai7i and ||F|| = y/Yrf.
2 = 1
6.53 Show that Ker(S) = JT*.
Hint: For m = 0,1, 2,..., define for j G {0,..., 2m - 1}
00 2n-l 00
PmJ [J2 Yl *(n,0e("i0j = Yl 12 V,0e(«,0
n=0 z=0 n~m(rn)j)<(n,i))
2m-l
and Pm = J^ Fmj. Then Pm are projections on JT. First, show that, for
x* G Ker(iS), lim ( max IIP* ,-(e*)||) = 0. If it were not the case, there
would be a sequence {(n^, ik)}keN and e > 0 such that \\P?nk ik)(x*)\\ > ?•
One can show that there exists only a limited number of incomparable
elements in {(njfe, 2jfe)}fceN, and thus there exists a branch 7 G T
containing infinitely many elements from {(n^z^J^N- We may assume that
{(n*,»*)}*€N C T and U^,^*) - P{nk+uik+l)(x*)\\ > e for all k.
By the previous exercise, we have P7(#*) G span{/(n>2); (n,i) G 7} and,
for k large enough, ||(f?fc - P*k+1)P7(x*)\\ < \e. Define for k G N maps
/7* — P* — P* (P* P* }P*
k r(nk)ik) r(nk+i,ik+1) \rnk rnk+1)r-y
Then ?/? is a dual projection to some Uk and ||?/?(z*)|| > \s. The
supports of the subspaces Uk(JT) are mutually disjoint and no branch y' G
T intersects more than one of them. Thus
,,jfe=i
and for j large enough we obtain a contradiction.
j
2 3
k = l
12

6. Schauder Bases 201
Now assume that there is v* G Ker(S') \ JT*, and let dist(v*, JT*) >
A — 6)||v* || for some 6. To get a contradiction, we find x, y ? Sjt such that
v(x) > 1 — 6, v(y) > 1 — 6, and Pmi(x) = 0, Pm2(y) — V f°r some m2 much
larger than mi. For 6 small enough, we use our limit formula and obtain
by calculating the norm of x-\-y from definition that ||a? + y|| < 3.5. Details
may be found in [LiSt].
6.54 Show that JT is a separable space with nonseparable dual containing
no isomorphic copy of l\.
Hint: Using Dixmier's projection, from the preceding exercise we obtain
that JT** - JT®l2{T). Thus card(JT**) = c. If there was a copy of lx
in JT, we would find a copy of l\{c) in JT* by Exercise 5.51 and thus
^oo(c) would be isomorphic to some quotient of JT**. This would imply
that card(JT**) > 2°, which is a contradiction.

7
Compact Operators on Banach Spaces
Definition 7.1
Let X and Y be Banach spaces.
T G B(X,Y) is called a compact operator if T(Bx) is compact in Y.
The space of all compact operators from X into Y with the norm inherited
from B(X,Y) is denoted by JC(X, Y). IfX = Y, then we write IC(X) instead
ofK(X,X).
T G B(X,Y) is called a finite-rank operator or a finite-dimensional
operator 2/dim(T(X)) < oo.
By ^(XjY) we denote the space of all finite-dimensional operators from X
into Y with the norm inherited from B{X)Y).
If / G Sx* and / does not attain its norm on Bx, then f(Bx) = (—1,1).
Thus we have / G /C(X, R), yet f(Bx) is not compact. Therefore, the
closure T(Bx) in the definition of a compact operator cannot be dropped.
Let T: X —> Y be a finite-rank operator. Consider some Auerbach basis
{et-;#} of T(X). We define /,- = g{ o T G X\ Then, for every x G X, we
have T(x) = Y,9iT{x)ei = Y,fi{x)ei- Conversely, given /i,...,/n G X*
and ei,..., en G Y, then T: z i—> Ylfi{x)ei ls a finite-rank operator from
X into Y. Such an operator is denoted by T — ^ /t- 0 e,-.
Proposition 7.2
id X, Y 6e Banach spaces. Then T(X,Y) is a subspace of K(X,Y).
JC(X,Y) is a closed subspace of B(X,Y), and hence a Banach space.

204 7. Compact Operators on Banach Spaces
Proof: Since (Tx + T2)(X) C Ti(X) + T2(X), T(X,Y) is a subspace of
B(X, Y). If T is a finite-rank operator, then T(Bx) is a bounded set in a
finite-dimensional (closed) space T(X), and hence T{BX) is compact.
For operators T\, T2 we also have
(aTi + CT2)(BX) C aTx{Bx) + PT2(BX) C aT^Bx) + f3T2(Bx),
and if 7} are compact, the right-hand side is a compact set (Exercise 1.55).
Thus /C(X, Y) is a subspace of B{X, Y). We will show that it is closed.
Consider Tn G K{X, Y) such that lim(rn) = T in B(X, Y). To show that
T is a compact operator, given e > 0, we find a finite ?-net for T{BX). First,
note that Tn -> T in #(X, Y) means that lim (Tn(z)) = T(x) uniformly
n—>-oo
for a? G 5x- Thus, there exists no such that ||Tn(x) — T(x)\\ < e/2 for
x G Bx and n > no. Since Tno(Bx) is totally bounded in Y, there is a
finite e/2-net F for TnoEx). We claim that F is a finite e-net for T(J3x).
Indeed, given x G #x, we find y G F such that ||jTno(z) - y|| < e/2. Then
l|r(x)-y|| < ||T(a?)-rno(x)|| + ||rno(x)-y|| <e. Therefore, T is a compact
operator.
D
Note that if X is infinite-dimensional, then no isomorphism into from
X into Y can be a compact operator by Theorem 1.24. In particular, the
identity operator Ix on an infinite-dimensional Banach space X is never
compact.
The spaces K{X,Y)* and IC(X,Y)** are discussed in [Mup].
Example
Define T G B(?2) by T(x) = B-,"a?i) for a? = (Xi). Then T G ?(*2) \ ^D0.
Indeed, since T(Bi2) is a closed subset of the Hilbert cube (Exercise 1.49),
it is compact.
Define T G B(?2) by T(x) = (a?i,0,ar3,0,...) for x = (a:,-). Then T is
a bounded linear operator because ||T(a?)[| = (^2\x2i\2J < (X^l^'l2)^-
However, T is not compact since {e2n+i} is a sequence in T(Bi2) that has
no convergent subsequence (note that ||e2»+i — e2y.fi
Lemma 7.3
Let X,Y be Banach spaces and T,Ti,T2,... G B(X,Y). If lim(Tn(z)) =
T{x) for every x G X, then for every compact set K in X we have Tn(x) —+
T(x) uniformly on K.
Proof: By contradiction, assume that there is a compact set K in X,
e > 0, a subsequence of {Tn} (denoted by {Tn} again) and xn G K such
that ||Tn(?n) — T(xn)|| > e. Since {xn} C K, we may assume that xn —> x
for some x G K. Note that by the uniform boundedness principle we have
M = sup{||T||, ||Ti||, ||T2||,...} < 00. Thus
\\(Tn-T)(xn)\\ < |j(Tn - T)(x)|| + ||(Tn - T)(x„ - x)[|
< \\(Tn - dooii + ||rn - r|| • \\xn - x\\

7. Compact Operators on Banach Spaces 205
< ||(Tn-T)(aOII + 2M.||zn-z||-+0,
a contradiction with ||Tn(zn) — T(zn)|| > e.
D
Proposition 7.4
Let X be a Banach space with a Schauder basis. Then T{X) — K,(X).
Proof: Let Pn be the canonical projection associated with the Schauder
basis {e2}. For every x G X, we have lim(Pn(x)) = x — Ix(x), where
Ix is the identity operator on X. Given T G IC(X), we claim that the
finite-dimensional operators Pn o T converge to T in B(X). To see this,
we must show that (Pn — Ix)(T(x)) converges uniformly to zero on Bx\
that is, (Pn — Ix) converges uniformly to zero on T(Bx)- This follows from
Lemma 7.3 because T(Bx) is compact.
?
Not all spaces have this property. In general, since the space of compact
operators is closed, we get T(X, Y) C IC(X,Y) for Banach spaces X,Y.
A Banach space Y is said to have the approximation property (A.P.) if for
every Banach space X we have T{X, Y) = /C(X, Y). A slight modification
of the preceding proof shows that c0 and ?p, p G [1, oo), have A.P.
Proposition 7.5
Let X be a Banach space with a Schauder basis. If X* is separable, then
)C(X) is separable.
Proof: First, we will show that one-dimensional operators form a
separable subset in JC(X). Choose a countable dense set {fi} in X* and a
countable dense set {#n} in X. Then the sequence of operators Titn:x h->
fi{x)xn is dense in the set of one-dimensional operators on X. Indeed, let
T be a nontrivial one-dimensional operator on X of the form T{x) = f(x)e,
where / G X*, e G X. Given e > 0, choose fi such that ||/ — /2-|| < ?/||e||
and xn such that ||e — xn\\ < ?/(||/|| + ?/||e||)- For \\x\\ < 1, we have
||/(«)e -/,-(i)i„|| < IWI-ll/-/ill-IWI + IIAII-ll«-*»ll-lkll
Thus HT^-T||<2?.
Since the span of one-dimensional operators is T{X)) this space is
separable. From IC(X) = J-(X), we get that JC(X) is separable.
D
Recall that #(^2) is not separable (Proposition 1.28).
Proposition 7.6
Let X,Y be Banach spaces and T G JC(XyY). If X, then
T(xn) -* T(x) in Y.

206 7. Compact Operators on Banach Spaces
An operator satisfying the conclusion of Proposition 7.6 is called a
completely continuous operator. Thus, every compact operator is completely
continuous. The example of the identity operator on ?i shows that the
converse implication is not valid in general.
Proof: If xn —> #, then {xn} is weakly and hence norm bounded, and we
may assume that x,xn G Bx> We have T(xn) —> T(x) by w-u>-continuity
of T. However, T(Bx) is a compact space in the norm topology, and the
u>-topology is weaker and Hausdorff; hence these two topologies coincide
on T(Bx)- Consequently, T(xn) —> T(x).
U
Example
Let X - L2[0,1] and K G L2([0,1] x [0,1]). Define an operator T from
Z,2[0,1] intoL2[0,l]by
T(x):t>~* / K(t,s)x(s)ds.
Jo
Note that, indeed, T(x) G L2[0,1] whenever x G L2[0,1]:
\\T(x)\\l2 = (J \J K(8,t)x(s)ds\2dt)i
al pi pi 1_
(/ \K(s,t)\2ds \x{s)\2ds") dty
= (/ x2(S)dSy([ I \K(S,t)\2dsdty < CO.
Thus also T G B(L2[0,1]) and ||T|| < (j^1 |#(M)|2 <***)*¦
We will show that T is a compact operator. First, we will show that if
K is continuous on [0,1] x [0,1], then T maps L2[0,1] into C[0,1]. By the
continuity of K, we have M = sup {\K(s,t)\\ (s,t) G [0,1] x [0,1]} < oo
and hence for x G Bl2 we get
\T(x)(t)\ = 1/ #(*, «)*(*) cfel < / |#(*, s)| \x(s)\ ds
Uo I Jo
< [J \K(t,s)\2dsy\\x\\L2<(j M2ds)h\\x\\L2<M.
Given e > 0, find <S > 0 (from the uniform continuity of K on [0,1] x [0,1])
such that if ti,t2 G [0,1], \t2 — t2\ < 6, then for every s G [0,1] we have
\K(ti,s) — K(t2,s)\ < e. Consequently, for every x G Bl2 and |/i —12\ < 6,

7. Compact Operators on Banach Spaces 207
we have
|r(s)(ti)-r(x)(t2)| = \J K(h,s)x(s)ds-J K(t2ls)x(s)ds
< [ \K(tus)-K(t2,s)\\x(s)\ds
Jo
< (Kj1\K(t1,s)-K(t2,s)\2ds)i\\x\\L2
< (J^2)'\\x\\La<e.
Therefore, if K is continuous on [0,1] x [0,1], then T(x) E C[0,1]. In fact,
we also showed that T(Bl2) is a uniformly bounded (by M) and equicontin-
uous set C[0,1]; hence, by the Arzela-Ascoli theorem, T(Bl2) is relatively
compact in C[0,1]. Since the norm topology of L2\QI] is weaker than the
topology of C[0,1], we have that T(Bl2) is compact in L2[0,1] and T is a
compact operator.
If K E ?2([0,1] x [0,1]), then choose a sequence Kn of continuous
functions on [0,1] x [0,1] such that
( (K(t,s)-Kn(t,s)J dsdtY -+0 as n ^ oo.
For n E N, define a compact operator Tn: L2[0,1] —> L2[0,1] by
T„(x):ti-> / Kn(t,s)x(s)ds.
Jo
By the first estimate of this example,
||T-Tn||L2 < (/ / \K(t,s)-Kn(t,s)\2dsdty ->0asn->oo.
Thus T E /C(L2[0,1]) = /C(L2[0,1]).
Theorem 7.7 (Schauder)
Let X,Y be Banach spaces and T E B{X,Y). T* E /C(Y*,X*) z/and <mfy
i/ TeK{X,Y).
Proof: Let T E K(X,Y). We must show that T*(BY*) is totally bounded
in X*. Let {/n} C 5y* be an arbitrary sequence.
Consider fn restricted to T(Bx), which is compact in Y. Then {/n} are
uniformly bounded and equicontinuous. By the Arzela-Ascoli theorem, the
restrictions of fn to T(Bx) form a totally bounded set in C(T(Bx)) • Hence
there is a subsequence fnk such that sup \fnk (T(x)) — fni (T(x)) | —>¦ 0 as
xeBx
kj —+ oo. Consequently,
lira ||T*(/nJ-T*(/ni)|| = lim sup \{T*(fnk)-T*(fni))(x)\

208 7. Compact Operators on Banach Spaces
= lim sup |/nt (T(x)) - fn, (T(xj) | - 0.
kfl-*oo xeBx
Thus T*(fnk) is Cauchy in X*, and T*(J3y*) is compact.
To prove the opposite implication, recall that T**|x = T. By the first
part, T* G /C(Y*,X*) implies that T**(BX») is compact. Since T**(BX)
is a closed subset of T**(Bx**), we get that T**(Bx) is compact in X**
and hence in X. Thus T G /C(X, Y).
?
Lemma 7.8
Ze2 X be a Banach space. Let T G B(X); denote S ~ Ix —T and Y — S(X).
IfY is a proper closed subspace of X, then for every e > 0 there is xq G Bx
such that dist(T(a?0),r(Y)) >l-e.
Proof: By Riesz's lemma (Lemma 1.23), there is xq G Sx such that
dist(x0,Y) > 1 - e. We have S(x0) G Y and T{Y) = (Ix - S)(Y) C Y.
Therefore, dist(T(z0),T(Y)) > dist(T(^0) + S(x0),Y) = dist(z0,Y) >
a
Theorem 7.9
Let X be a Banach space. Assume that T G fc(X) and A =^ 0. Then
Ker(Aix — T) is finite-dimensional, and (XIx — T)(X) is closed and
finite-co dimensional.
Proof: We may assume that A = 1. Let N\ = Ker(Ix — T). For every
x G N\, we have T(x) = x, and hence T\ is an isomorphism into and
also compact, so N\ is finite-dimensional.
By Theorem 5.6 and Proposition 5.3, there is a closed subspace X\ oiX
such that X - N\ 0 X\. Denote S = Ix - T, S\ = S\x , and note that
S(X) = 5(Xi) = Si(Xi). Since Ker(Si) = iVA (\X1 = {0}, we get that Si
is one-to-one. We will show that inf ||5i(a?)|| > 0.
x?SXl
By contradiction, assume that there are Xfi G Sxr such that ||Si(#n)|| —»
0. Since T is compact, we may assume that T(xn) —¦» y. Then xn — (Si +
T)(ar„) —> y. Therefore, ||y|| = 1 and also Si(zn) —> S\(y), so Si(y) = 0.
This contradicts Si being one-to-one.
Thus, there is c > 0 such that ||Si(x)|| > c\\x\\ for all x G X\ and, by
Exercise 1.27, S\{X\) — S(X) is closed.
We will now prove that S(X) is finite-codimensional. For k G No, define
Sk as S° = IX, S1 = S, Sk+1 = S o Sk. Let Nk = Kei(Sk). Since Sk =
(J* — T)fc = Ix — Tk for some compact operator Tfc (powers of T are
again compact operators), we have dim(Nk) < co for every k. Denote
Mk = Sk(X) = Sfc(Xi). We have N0 C Ni C N2 ¦ ¦ -, and Af0 D Mi • • •. We
claim that there is n such that Mn = Mn+i. By contradiction, if all the
inclusions Mq D Mi D ... were strict, we could find by Lemma 7.8 applied

7. Compact Operators on Banach Spaces 209
to S\Mj.Mn -+ Mn elements yn G BMn such that dist(T(yn),T(Mn+i)) >
\. This would in particular mean that ||T(yn) — T(ym)|| > \ for n / m, a
contradiction with the compactness of T.
Similarly, there is m such that Nm = Nm+i. Indeed, if x G Nk (i.e.,
Sk(x) = 0), then S*^*)) = 0 and thus S(x) G Wfc_i C Nk.
Therefore, we again use Lemma 7.8 for S\N : Nj~ —* Nk to obtain the claim.
Consequently, Mn = Mn> for any n' > n and Nm — Nmi for any mf >m.
Finally, we claim that for p — max{n,ra} we have X — Np 0 Mp. For
arbitrary x G X, we have SP(s) G Mp. However, SP(MP) = ^(^(X)) -
52p(X) = 5P(X) = Mp. Therefore, there exists some y E Mp such that
5p(y) = SP(s), so Sp(y - x) = 0. Thus y - a: G Np and x = (x - y) + y.
Since X = 7Vp © Mp, the codimension of Mp (and hence of Mi D Mp) is
finite, and the proof is complete.
?
In particular, (XIX-T)(X) = Ker(AIx* -T*)± (see Exercise 2.34). It can
be proved that (AIX* -T*)(X*) is w*-closed, and hence (XIX* -T*)(X*) =
Ker(AJx - TI (c/., Exercise 3.37).
Definition 7.10
Let X, Y be Banach spaces. An operator T G B(X) Y) is called a Fredholm
operator if Ker(T) is finite- dimensional and T(X) is finite-co dimensional.
The number i(T) = dim(Ker(T)) - codim(T(X)) is called the index ofT.
As in the proof of Theorem 7.9, if T is a Fredholm operator, we can write
X — Ker(T) 0 Xi, and T\ is an isomorphism of X\ onto T(X).
From Theorem 7.9, we immediately obtain the following proposition.
Proposition 7.11
Let X be a Banach space and T G IC(X). Then XIx — T is a Fredholm
operator for every X / 0.
Example
Let X = Y = l2. For k G N, define Tfa) = (xi+k) G I*. Then Ker(T) =
span{ei,... ,e^}, Xi = span{efc+i,...}, T(X) — X, and T is a Fredholm
operator with i(T) = k.
Theorem 7.12 (Fredholm alternative)
Let X be a Banach space, and let T G fc{X) and X ^ 0. Then the equation
T(x) — Xx — y has a solution for every y G X if and only if the equation
T(x) — Xx — 0 has only the trivial solution x = 0.
In other words, Ker(A/x - T) = {0} if and only if (XIX - T)(X) = X.
In fact, a more general result is true: If T is a compact operator on X and
A ^ 0, then i(XIx - T) = 0 ([Mup]).
Recall that for 5 G B{X) we have codim(S(X)) = dim(Ker(S*)) if they
are both finite.

210 7. Compact Operators on Banach Spaces
Proof: We can assume that A = 1; denote S = Ix - T. If T(x) - x = 0
has only the trivial solution x = 0, then 7V~a = Ker(S) = {0} and thus S is
an isomorphism into by the proof of Theorem 7.9. We must show that it is
in fact onto.
Set Mk = Sk(X) for k — 0,1,.... In Theorem 7.9, we proved that there is
n so that Mm = Mn for all m> n. We claim that Mi = M0 = X. If this is
not the case, let m be the smallest integer such that Mm_i / Mm = Mm+i.
Pick u E Mm_i \ Mm. Then S(u) E Mm = Mm+i. Therefore, there is
v E Mm such that S(t>) = S(u) and w ^? t; since u ? Mm. Hence S(u—v) = 0
and u ^ v, a contradiction with Ker(S) = {0}.
Now assume that S maps X onto X. Define Nk — KerEfc) for k E N.
We must show that Ni = Ker(S) - {0}. Clearly, Nk C Nk+i for every
fe. Assume by contradiction that there is x\ ^ 0 such that x\ E N\. By
induction, we will construct a sequence xk such that S(xk+i) = a^jb and
^ib G Nk \ Nk-i. This will complete the proof, since we know from the
proof of Theorem 7.9 that Nm = Nm+i for some m.
Assume that x\,..., xk were constructed. Since S is onto, there is xk+\
such that 5(a?jb+i) = xk. Then 5*(a?jfc+i) = Sk~l{xk) — ... = x\ / 0 and
5A;+1(xjfc+i) = 5(^1) = 0, which concludes the proof.
?
Spectral Theory
Definition 7.13
Let X, Y be Banach spaces. An operator T E B{X) Y) is called invertible if
T is an isomorphism of X onto Y.
T E B(X, Y) is invertible if and only if there is a bounded linear operator
T E B(Y, X) such that T~lT = Ix (the identity on X) and TT~l = Jy.
By the open mapping theorem, this is equivalent to T being one-to-one and
onto.
Thus, T E B{X, Y) is invertible if and only if T* is invertible, and
(T*) = (T)*. Also, if T E B{X,Y) and S E B(y, Z) are invertible,
then ST is invertible and (ST)-1 =T-1S~1.
Lemma 7.14
Let X be a Banach space and T E B(X). If \\T\\ < 1, then (Ix - T) is
oo
invertible and (Ix — T)-1 = Yl Tk, where the series converges absolutely
k=0
in B(X).

7. Compact Operators on Banach Spaces 211
Proof: First, note that ? ||r*|| < ? ||T||* = jzwjr, so X] Tk is
fc=0 fc=0 fc=0
absolutely convergent in B(X). Hence
(Ix-T)Y^Tk = (Ix-T) + (T-T2) + -.- = Ix.
@0 \
ET*)(Jx-r) = jjf.
k=0 '
a
Lemma 7.15
Zei X be a Banach space and S,T E B(X). IfT is invertible and \\S — T\\ <
IIT!!-1, then S is invertible and
"* N-i-||r-i||||5-r||-
Proof: We have [{T-^T - S)\\ < [|r—1|| • ||T - 5|| < 1. Therefore, by
Lemma 7.14,1—T~1(T—S) = T-15 is invertible, and hence 5 is invertible.
00
We also have [Ix - T~\T - S))~l = EOT^ " S)T'• Hence
*=o
oo
S = (T-(T-S))-1 = (T^-r-^T-S)))-1 = ^(T-^T-S))^-1
n=0
and thus
CO
\\S-l~T~l\\ < ^WiT-^T-S^T^W
n = l
oo
lir-Tlir-SH
n=l
D
Corollary 7.16
Let X be a Banach space. The set C of all invertible operators on X is an
open set in B(X) and the map T *-+ T~l is a homeomorphism of C onto C.
Definition 7.17
Let X be a Banach space over K, T G B(X). The spectrum cr(T) ofT is
defined by
a(T) = {A G K; XIX - T is not invertible}.
The resolvent set p(T) is defined by p(T) = K \ <r(T). The points of p(T)
are called regular values of T.
If^t p(T), then R(X) = (XIX - T)~l is called the resolvent ofT at X.

212 7. Compact Operators on Banach Spaces
Note that if X is infinite-dimensional and T ? /C(X), then 0 ? a(T)
because otherwise T would be a compact isomorphism.
Fact 7.18
Let X be a Banach space and T ? B(X). Then a(T) = <r(T*).
Proof: (XI — T) is invertible if and only if (XI — T)* is invertible, and
(XIX-T)*=XIX.-T*.
?
Proposition 7.19
Let X be a Banach space over K and T ? B(X). The spectrum c(T) ofT
is a compact set in K bounded by \\T\\.
Proof: By Corollary 7.16, it follows that p(T) is an open set in C and
thus a(T) is a closed set in K. If |A| > ||T||, then (XIX -T)~l = X-l(lx -
_1 oo rpn
x) = S T^TT exists by Lemma 7.14 and A g <r(T). Thus, a(T) is
n=0 An+
bounded (by ||T||) and closed, and hence compact.
?
The next two statements are valid only in the complex case. We will use
several results from the theory of complex functions. Recall that if Z is
a complex Banach space and D is an open set in the complex plane C,
a function f:D-^Zis said to be analytic if for every zq ? D there is
oo
r = r(zo) and an ? Z such that f(z) = Y2 an(z — zq)u for z ? D(zo, r) =
n=0
{z ? C; \z — zo\ < r} C D and the series is absolutely convergent in
D(z0ir).
We will now show that Liouville's theorem remains valid for complex
Banach space-valued analytic functions. Precisely, if Z is a complex Banach
space and /: C —> Z is an entire function such that sup ||/(z)|| < oo,
then / is a constant function. Indeed, if h ? X*, then g(z) = h(f(z)) is
an entire function on C since h(^an(z — z0)n) = ^/i(an)(z — z0)n and
J2\h(an)\(z-zo)n < IHIEIIan||(*-2o)n < oo. Therefore, by the standard
Liouville theorem, g is constant on C (i.e., g(z) = #@) for every z ? C).
Hence h(f(z) - /@)) = 0 for all h ? X\ and thus f(z) = /@).
For other results used in the proofs of the following theorems, we refer
to [Rud2].
Theorem 7.20
Let X be a complex Banach space. For every T ? B(X), we have c(T) / 0.

7. Compact Operators on Banach Spaces 213
Proof: Fix A0 G p(T) and choose A satisfying |A0-A| < [[(Ao/^ —T)_1 H.
By Lemma 7.15 applied to Xolx — T and XIx — T, we get
oo
R(X) = (A/x-^-^^po/x-^-^Ao-A^rCAo/x-r)-1
oo oo
= ?(A0 - A)"(A0/a- - T)-n~l = ?(A0 - A)n/2(A0)n+\
n=0 n=0
where the series is absolutely convergent. We have just proved that the
resolvent function R is an analytic function on p(T) with values in the
Banach space B(X).
In the proof of Proposition 7.19, we observed that for |A| > ||T|| we
have i?(A) = ?^ ^rri hence ||iJ(A)|| < rxpiTw- In particular, R(X) —> 0 as
|A| —> oo. Therefore, assuming p(T) = C, by Liouville's theorem we would
have that R = 0 on C, which is impossible because R(X) is an inverse
operator. Hence <r(T) ^ 0.
?
Definition 7.21
Let X be a Banach space and T G B(X). The spectral radius r(T) ofT is
defined by r(T) = sup{|A|; A G <r(T)).
By Proposition 7.19, we have r(T) < \\T\\.
Theorem 7.22 (Gelfand)
Let X be a complex Banach space. For every T G B{X), we have
KT) = lim(||Tli).
In the proof, we will use the following statements.
Fact 7.23
Let X be a Banach space. Assume that T,S G B(X) commute; that is,
TS = ST. Then ST is invertible if and only if both T and S are invertible.
PROOF: We already observed that if T and S are invertible, then so is ST.
On the other hand, assume that ST is invertible. We claim that T and S
are both one-to-one. Indeed, if for x / 0 we have T(x) — 0, then (ST)(x) —
5@) = 0 and ST is not one-to-one. If for some x ^ 0 we have S(x) = 0,
then TS is not one-to-one again. If T(X) ^ X, then TS is not onto and
similarly we proceed if S(X) =fc X, using the commutability of S and T.
Thus, T and S are one-to-one and onto, and hence invertible.
?
Fact 7.24
Let X be a complex Banach space. IfT is an operator in B{X) and n G N,
thena(Tn) = {/in; pe<r(T)}.

214 7. Compact Operators on Banach Spaces
Proof: For A E C, factor the complex polynomial tn — A as (t — Ai) •
(t - A2) • • • (t - A„) for every t E C. Then (Tn - XIX) = (T - XJX) •
(T — \2lx)'' - {T — \nIx)- By the inductive use of Fact 7.23, we get that
Tn — XIx is invertible if and only if (T — X{Ix) are invertible for all i. This
means that A E o~(Tn) if and only if at least one n-th root A; of A is in
a(T). Thus, A E cr(Tn) if and only if A = /in for some /i E <r(T).
D
Proof of Theorem 7.22: (Sketch) By Fact 7.24, a(Tn) = {*n; t E
cr(T)}, so r(Tn) = r(T)n. By Proposition 7.19, we have r(Tn) < ||rn||,
and therefore r(T)n < \\Tn\\ for every n E N. Hence r(T) < ||Tn||« for all
n and thus r(T) < liminf ||Tn||» .
On the other hand, we proved that R(X) is an analytic function on p(T).
For |A| > ||T||, we have an expansion R(X) = j ^Z x^- ^y ^ne properties
of Laurent series, this expansion converges for all A such that |A| > r(T),
so there is C such that ^||^r|| < C. These facts follow from the general
n
theory of Banach space-valued analytic functions ([Rud2]) and are by no
means obvious.
Thus \\Tn\\n < \X\Cn -> |A|. Since |A| > r{T) was arbitrary, we get
r(t) < liminf ||Tn||i < limsup ||Tn||^ < r(t), so lim||Tn||^ = r(t).
D
Considering the operator of rotation by |- T(K(xi)X2J) — (—^23^i) on
R2, we see that Theorem 7.20 and Theorem 7.22 are not valid in the real
case.
Definition 7.25
Let X be a Banach space and T E B{X). A scalar X is called an eigenvalue
of T if there is 0 ^ x E Ker(AZx — T). Such x is called an eigenvector
corresponding to the eigenvalue X. The subspace Ker(AIx — T) is called the
eigenspace corresponding to the eigenvalue A.
Every eigenvalue A of T lies in its spectrum, since XIx — T cannot be
invertible. If dim(X) < oo, then the operator (XIx — T) is invertible if and
only if it is one-to-one. In this case, o~(T) is equal to the set of all eigenvalues
of T. However, in general not every point of cr(T) is an eigenvalue.
Lemma 7.26
Let X be a Banach space, T E B(X). Assume that Ai,..., An are distinct
eigenvalues ofT. If ei is an eigenvector corresponding to A; fori = 1,..., n,
then {ei, ...,en} is linearly independent.
Proof: By induction. Assume that ei,..., en_i are linearly independent,
n — 1 n —1 n —1
and let en — J2 azef. Then J2 ^n&iZi — Xnen = T(en) = ]T AzQf2ez-; that
2 = 1 8 = 1 8 = 1

7. Compact Operators on Banach Spaces 215
n-l
is, Y2 fin — A2)a2ez- = 0. Since ei,... ,en_i are linearly independent and
t=i
An — Xi ^ 0, we get a* = 0 for all i.
?
Let X be a Banach space and T ? S(X). A closed subspace Y of X is
called invariant for T if T(Y) C Y. Obviously, {0}, X, and all eigenspaces
of T are invariant for T.
It is not known whether every bounded operator on a Hilbert space has
an invariant subspace other than {0} and the whole space (a
nontrivialinvariant subspace). However, Enflo constructed the first example of a Banach
space X and an operator from B(X) without a nontrivial invariant sub-
space (see, e.g., [Bol]). Note that there exist compact operators without any
eigenvalue (see the following Example). It is known, in contrast, that every
compact operator has a nontrivial invariant subspace; see Theorem 7.62.
Example
Consider the operator T ? B(L>2[0,1]) (over the complex scalars) defined
by T(x):t^tx(t). Clearly, ||T|| < 1.
We claim that T has no eigenvalues. Indeed, if A ? C and for some x(t)
from Li we have \x(t) — tx(t) = 0, then (X — t)x(t) = 0 almost everywhere
on [0,1], so considering f/Awe get x(i) — 0 a.e.
However, we will now show that [0,1] C &(T) (so in particular ||T|| = 1).
Fix A ? [0,1] and choose e > 0 such that [A, A + e] C [0,1] or [A - e, A] C
[0,1]. Assume that [A, A + e] C [0,1] is the case and put
^ for t ? [A,A + 6r]
0 for 2 <? [A,A + ?]
(the definition of x? in the case [A — ?, A] C [0,1] is analogous). We easily
observe that fQ x^(t)dt = fx ? j = 1, so x? ? Sl2- On ^ne other hand,
(XIx — T)(xe):t\-*(X — /)?e(tf), and therefore
\\(\ix - T)(xs)\\2 = fX+e -a - tf dt = - fX+\\ - tf dt
J\ e e JX
lr (X-tK]^x+e i e2
ei 3 Jt=A 3?
so (XIx —T)(x?)^0 as e —* 0. Therefore, (XIx — T) is not invertible.
It is easy to observe that the operator T has a rich structure of non-
trivial invariant subspaces. For example, the subspace I/2[0,r] of i^fO, 1] is
invariant for every r ? [0,1].
Proposition 7.27
Let X be a Banach space and T ? K(X). If X ? cr(T) \ {0}; then X is an
eigenvalue ofT.

216 7. Compact Operators on Banach Spaces
Proof: If A ^ 0 is not an eigenvalue, then Ker(AIx - T) — {0}. By
Theorem 7.12, XIx — T is also onto, and hence invertible. Thus A ? 0*(T).
?
Proposition 7.28 (Riesz, Schauder)
Let X be a Banach space, T G K(X). For every e > 0, T has only finitely
many eigenvalues with absolute value larger than e.
Proof: Assume that there is an infinite sequence {A2} of distinct
eigenvalues such that |A;| > e for every i. For every A;, choose an eigenvector
xi. For n G N, define Xn = span{#i,..., xn}, and note that T(Xn) — Xn
and Xn-i / Xn by Lemma 7.26.
By Lemma 1.23, we obtain yn G Xn such that distB/n,Xn_i) > \ and
||t/n|| — 1 for every n. Put zn = yn/^n and note that ||zn|| < j. We have
n
T(zn) G Xn and also yn - T(zn) G Xn_i, since for yn = X] cfcZA; we have
jfe=i
n n —1
yn - T{zn) = ^A - ?)<*** = 2A " ^)C*** € Xn_i.
fc = l Jb = l
If n > m, then T(zm) G Xm C Xn_i and yn - T(zn) G Xn_i, and thus we
have
||T(zn)-r(zm)|| > dist^^),*^)
= dist(T(sn) + yn - r(z„),X„-i)
= dist(s/n,Xn_i) > |.
We obtained an infinite bounded sequence {zn} such that [[T^n) —
^Xzm)|| > \ for m ^ n, a contradiction with the compactness of T(Bx).
?
Theorem 7.9 together with Proposition 7.28 yields the following
statement.
Corollary 7.29
Let X be a Banach space and T G K(X). Then cr(T) = {0, Ai, A2,...},
where {Xi} is either a finite set (possibly empty) or a sequence tending to
zero, formed by nonzero eigenvalues, each of Xi having a finite-dimensional
eigenspace.
Self-Adjoint Operators
Let H be a Hilbert space and T G B(H). Using Theorem 2.21, we
observe that there exists a unique operator Q G B(H) satisfying (T(#),y) =
(x,Q(y)) for every x,y E H.

7. Compact Operators on Banach Spaces 217
Definition 7.30
Let T be an operator on a Hilbert space H. The adjoint operator to T,
denoted T*, is defined by
(T(x),y) = (x,T*(yj) forallx,yeH.
In order to preserve the customary notation, for the rest of this section
the symbol T* is reserved for the adjoint operator rather than for the usual
dual operator.
Let H be a Hilbert space, T E B(H), and let T* E B(H) be the adjoint
operator. We easily observe that T** = T. Note also that I*H = IH, (AT)* =
AT*, (ST)* = T*S\ and (T + 5)* = T* + 5* for T, 5 E B(JT).
Denote for the moment by Td the dual operator Td E B(H*). Let /, # E
i7*, and let a, 6 E H be vectors assigned to /, g by the Riesz identification
(Theorem 2.21); that is, f(x) = (xya) and g(x) = (a:, 6) for all x ? H.
It follows from our definitions that Td(f) = g if and only if T*(a) = 6.
Similarly to Fact 7.18, we prove the following fact.
Fact 7.31
Let H be a Hilbert space, T E B(H). A E <r(T) if and only if A E <r(T*).
Proposition 7.32
Ze^ fl" 6e a jyifter/ space anrf T E B^). TAen Ker(T) = T*(H)L.
Proof:
Ker(T) = {z E #; T(a?) = 0} = {xGiJ; (T(x), y) = 0 for all y ? H}
= {xeH; (s,T*(y)) = 0 for all y E ff} = T*^)^.
?
In particular, if = Ker(T) 0 T*(H).
Definition 7.33
Let H be a Hilbert space. An operator T E B(H) is called self-adjoint or
hermitian if T* = T; that is, (T(x),y) = (x,T(y)) for allx,yeH.
Note that TT* and T*T are self-adjoint for any T E B(H).
Proposition 7.34
Let H be a complex Hilbert space. For every T E B(H), there are self-adjoint
operators Ti, T<i on H such that T = Ti+iT2. Moreover, this decomposition
is unique.
Proof: Put Tx = |(T-f T*), T2 = -\i(T-T*). Then Ti>2 are self-adjoint
and Ti + iT2 = T.
If Ti, T2 are self-adjoint operators such that Ti + iT2 — 0, then T\ — iT2 =
(T\ + iT2)* = 0. Adding and subtracting these two equations, we get T2 = 0
and Ti = 0, from which the uniqueness follows.
?

218 7. Compact Operators on Banach Spaces
Similarly to the polarization identities for a norm of a Hilbert space, in
a complex Hilbert space H we have for every T E B{H):
4(r(x),y) = (T{x + y),x + y)-(T(x-y),x-y)
+i(T(x + iy),x + iy) - i(T(x - it/), ar - iy) .
If H is a real Hilbert space and T is a self-adjoint operator on H, then
4(r(z), y) = (T(x + y), * + y) - (T(z - y), x - y).
Proposition 7.35
Zei H be a complex Hilbert space and T E B(H). If (T(x), a?) = 0 /or e^ery
zE#; thenT=:0.
Let H be a real Hilbert space. If T is self-adjoint and (T(x),x) — 0 for
every x E H, then T = 0.
Proof: By polarization identities for operators, we have (T(x)yy) = 0 for
every x,y E H\ in particular, (T(z),T(z)) = 0 for every x E H. Therefore
T=0.
D
For self-adjoint operators, this is a consequence of a more general result
to follow.
Example
Let T be the operator on R2 given by T((#i, x^)) — (—#2, #i) (rotation by
7r/2). Then (T(o;),a:) = (—0:2^1 + ^1^2) = 0 for x E R2, so Proposition 7.35
does not hold in general in the real case for operators that are not self-
adjoint.
Proposition 7.36
Let T be an operator on a Hilbert space H. IfT is self-adjoint, then
\\T\\= sup \(T(x),x)\.
xtBH
Proof: If \\x\\ < 1, then \(T(x),x)\ < \\T(x)\\ ¦ \\x\\ < \\T\\ • ||z||2 < ||T||.
Thus sup \(T(x),x)\ < \\T\\. Put sup \(T(x),x)\ = C. Note that
x€BH x€BH
\(T(z),z)\ < C\\zf for every zeH.
If z € H, z ± 0, put A = t11^^)* and u = \T{z). Using the fact that
(T(Xz), u) is a real number, and the parallelogram equality, we get
||T(,)||2 = (T(\z),j:T(z)) = (T(\z),u)
= ±[(T(\z + u),\z + u) - (T(\z-u),Xz-u)}
< \C{\\\z + W||2 + ||Az - U||2) = |C(||A.||2 + ||U||2)
= iC(A2|H|2 + ^||T(z)||2) = C||2||-||T(,)||.
Therefore, \\T(z)\\ < C\\z\\ for every z?ff, and hence ||T|| < C.
a

7. Compact Operators on Banach Spaces 219
Lemma 7.37
Let H be a Hilbert space H, T G B(H). IfT is self-adjoint, then (T(x),x)
is a real number for all x G H, and all eigenvalues ofT are real numbers.
Proof: Let x G H. Then (T(x),x) = (x,T(x)) = (T(x),x).
If A is an eigenvalue of T with an eigenvector x, then (T(x),x) =
(Xx, x) — X(x, x), and hence A = ' iijip ^s rea'-
Proposition 7.38
Let H be a Hilbert space and T G B(H). IfT is self-adjoint, then \\Tn\\ =
\\T\\n for every n > 1.
Proof: We can estimate
||T||2 = sup (T(x),T(x)) = sup (T*T(x),x)
x?BH x?BH
< ||r*T||<||r*||.||T||^||r||2.
Therefore \\T*T\\ = \\TT*\\ = \\T\\2. Since T is self-adjoint, ||T||2 = ||T2||.
The operator Tk is also self-adjoint, so we have ||T2 || — ||T||2 for every
*. If 1 < n< 2k, then
||T||2fc = ||T2fc|| = ||TnT2fc-n|| < ||Tn|| • \\T\\2k~n < ||T||n||T||2;c-n = ||T||2*;
hence ||T"|| • ||T||2fc-n = ||T||2*. Thus ||Tn|| = ||T||n.
?
Proposition 7.39
Let T be a self-adjoint operator on a Hilbert space H, and let X be a scalar.
Then X G <r(T) if and only if inf ||(AJjy - T)(x)\\ = 0.
x?SH
Proof: If A G p(T), then (XIH - T) G B(H) and for x G SH we have
1 = ii^n = iKAz^-^-^A/^-^x^H < (KA/^-^-^i.iKAJ^-r)^)!!.
Hence inf \\(XIH - T)(x)\\ > \\(XIH - T)~^.
||a:||=i
Now assume that inf \\(XIH-T)(x)\\-C > 0 for some C > 0. We then
x^Sh
have ||(A/# — T)(^)|| > C||^|| for every x G X, and hence XIh — T is an
isomorphism into (Exercise 1.27). In particular, its range is closed in H.
We will prove that XIh —T is also dense in H, thus showing that XIh — T
is invertible.
By contradiction, assume that (XIh — T)(H) is not dense in H. Then, by
Proposition 2.8 and Theorem 2.21, there is yo G ^, yo / 0, such that
((XIh - T)(x),y0) = 0 for every x G H. Since ((XIH - T)(x),yo) =
(x, (XIh —T)(yo)) for every x G H because T is a self-adjoint operator, we
get that (x, (XIH ~ T)(y0)) = 0 for every x e H; hence (XIH - T)(yQ) = 0

220 7. Compact Operators on Banach Spaces
and 2/0 i1 0. This means that A is an eigenvalue of T. Since all
eigenvalues of T are real, (XIh — T)(yo) = 0 and y0 ^ 0, contradicting
\\(XIH - T)(y0)\\ > C\\y0\\. Therefore, (XIH - T)(H) is dense in H.
D
Theorem 7.40
Let T be a self-adjoint operator on a Hilbert space H'. Define numbers
mT = inf (T(x),x) and MT = sup (T(x),x). TAerc <r(T) C [mT,MT]
x?SH x?Sh
(closed interval on the real line) and ttyit^Mt E &(T).
Proof: First, we will show that o~(T) lies on the real line. Assume that H
is a complex Hilbert space, and take any A = a + i/3. For every x G Sh , we
write
(As - T(x), a:) - (x, Xx - T(x)) = (A - A)(x, a:) = 2i/?;
hence
2|/?| = | (Xx - T(x), x) - (x, Xx - T(x)) \
< \(Xx-T(x),x)\ + \(x,Xx-T(x))\
< \\Xx - T(x)\\ • ||*|| + ||x|| • ||A* - T(x)\\ = 2\\(XIH - T)(x)\\.
This shows that inf \\(XIH - T)(x)\\ > |/?|, so A G cr(T) implies |/?| = 0
x?SH
by Proposition 7.39; that is, A is a real number.
Considering S — T + /xi#, we have a(S) — o~(T) -f//, ms — tut +//, and
Ms — Mt + //; so we may assume that 0 < my < Mt-
Since then ||T|| — Mt by Proposition 7.36 and cr(T) C R, we have that
<t(T) C [-Mt,Mt] by Proposition 7.19. We will show A = mT - d <? cr(T)
for every d > 0. For x G Sh , we estimate
((T - \IH)(x), x) = (T(x), x) - (Ax, a?) > mT - X\\x\\2 = mT - X = d.
Since also \((T-XIH)(x),x)\ < \\(XIH ~T)(x)\\\\x\\ = ||(A/X -T)(*)||, we
get inf \\(XIH - T)(x)\\ > d > 0. By Proposition 7.39, A g a(T) and we
established cr(T) C [mx,MT].
Now we will show that Mt 6 <r(T). Again, we will use Proposition 7.39.
Let xn G Sh be such that (T(xn), xn) —> Mt- Recall that we are assuming
0 < mT < MT\ hence Mt — ||T||, in particular, ||T(xn)|| < MT. Using the
fact that (T(a?n), xn) is real, we get
0 < \\{MTIH-T){xn)\\2 = \\MTxn-T(xn)\\2
= M$\\xn\\2 + \\T(xn)\\2 - 2MT(T(xn), xn)
< MZ + M%-2MT(T(xn),xn)^0.
Therefore \\(MTIH - T)(xn)\\ -* 0 and hence MT G <r(T).

7. Compact Operators on Banach Spaces 221
By considering S — T — MtIh, we get ms < Ms = 0, so \ms\ — \\S\\
and we prove in the same way that ms G &(S)] that is, mj G cr(T).
D
In particular, ||T|| G <r(T) if T is self-adjoint.
Lemma 7.41
Let T be an operator on a Hilbert space H. If T is self-adjoint, then
eigenvectors corresponding to different eigenvalues are orthogonal.
Proof: If T{x\) — \x\ and T(x2) — \ix2 for x\ / 0 and x2 ^ 0, A / //,
then (recalling that A,// must be real)
(T(xi),x2) = (Axi,ar2) = A(aJi,ar2)
(T(a?i),ar2) = (^i,T,(a?2)) = (a?i,/ia:2) = A*(si,a:2).
Therefore (A — fJ.)(xi,x2) = 0. Since (A — //) / 0, we have (#i, #2) = 0.
D
This implies that eigenspaces corresponding to distinct eigenvalues
of a self-adjoint operator are mutually orthogonal. We observed that
eigenspaces are invariant subspaces for the given operator. Recall that given
a closed subspace M of a Hilbert space #, its orthogonal complement ML
in H satisfies H = M 0M1.
Proposition 7.42
Let T be a self-adjoint operator on a Hilbert space H. Let M be a closed
subspace of H that is invariant under T. Then N — ML is invariant under
T. Denote T\ — T\ and T2 = T\N. Then Ti is a self-adjoint operator
on M, T2 is a self-adjoint operator on N, T{H) = T\(M) 0 T2(N), and
<r(T) = *(T1)U<r(T2).
Proof: Let y G N. Since M is invariant under T, for every x ? M we have
0 = (T(x),y) = (x,T(y)) and thus T(y) G N. Therefore, N is invariant
under T.
Since both M and N are invariant, the restrictions are self-adjoint
operators on the corresponding subspaces.
Let A G 0"(Ti). By Proposition 7.39, there are xn G Sm such that \\\xn —
Ti(xn)\\ -» 0. Therefore ||Aa?n - T(a?n)|| -+ 0, showing A G o-(T). Similarly,
we show that a(T2) C <r(T).
Assume that A ^ cr(Ti) U 0"(T2). Then there is C > 0 such that for
every x G M and for every y ? N, we have ||Aar — T(x)|| > C\\x\\ and
\\Xy - Ty\\ > C\\y\\. Write any z?H^z-x-\-y with x G M and ?/ G N.
Since Ax - T(ar) G M and Ay - T(y) G AT, we get
||Az-T(,)||2 = \\Xx-T(x)f + \\\y-T(y)\f > C*(||*||2 + ||y||2) = C2\\z\\\
Thus A g ct(T) by Proposition 7.39.
?

222 7. Compact Operators on Banach Spaces
We will now put together the results of the previous two sections.
Proposition 7.43
Every compact self-adjoint operator on a Hilbert space has an eigenvalue.
Proof: If T — 0, then 0 is an eigenvalue. If T ^ 0, then, by Theorem 7.40,
||T|| ? o-(T), and ||T|| ^ 0 is an eigenvalue by Proposition 7.27.
?
Theorem 7.44
Let T ^ 0 be a compact self-adjoint operator on an infinite-dimensional
Hilbert space H. Then <r(T) = {0} U {A,}, where Az- are distinct real
nonzero eigenvalues ofT. The set {At-} contains \\T\\ and is either finite or
a countable sequence convergent to zero.
Moreover, the space H has an orthonormal basis formed by eigenvectors
corresponding to eigenvalues ofT.
Proof: Since H is infinite-dimensional and T is compact, 0 ? &(T).
Since also ||T|| ? &(T), there is at least one nonzero eigenvalue, and by
Proposition 7.28 and Theorem 7.40 there is at most countably many real
eigenvalues ofT convergent to zero.
It remains to show that we can form an orthonormal basis out of
eigenvectors of T. For an eigenvalue A, denote N\ = Ker(A/# — T). We form
an orthonormal basis B\ of each N\. By Lemma 7.41, B — \jB\ is an
orthonormal set in H; clearly, span(jB) contains all eigenvectors of T. If
spanE) / H, consider G = spanEI. Since all eigenspaces are invariant
for T, so is spanE) and hence, by Proposition 7.42, G is also an invariant
subspace for G. Moreover, a(T) = flr(T|__ ) + <r(T\G). However, T\G
has an eigenvalue, and hence a nonzero eigenvector v. It must also be an
eigenvector ofT and thus v ? G C\ span(?), a contradiction. This shows
that spanE) = H\ hence B is an orthonormal basis of H.
?
Corollary 7.45
Let T be a compact self-adjoint operator on a Hilbert space H. Then cr(T)
is the closure of eigenvalues ofT.
Proof: If H is finite-dimensional, then every A ? o~(T) is an eigenvalue.
If H is infinite-dimensional, then the only point of <t(T) that need not be
an eigenvalue is 0. If the set of nonzero eigenvalues is countable, then it
converges to 0 and we are done.
The last case is that the set of nonzero eigenvalues is finite and H is
infinite-dimensional. Since eigenspaces of nonzero eigenvalues are finite-
dimensional (Theorem 7.9), the only possibility for eigenvectors to form an
orthonormal basis is that 0 is also an eigenvalue.
?

7. Compact Operators on Banach Spaces 223
Theorem 7.46 (Spectral decomposition)
Let T be a compact self-adjoint operator on an infinite-dimensional
separable Hilbert space H. Then there is an orthonormal basis {e2} of H such
that each e,- is an eigenvector corresponding to some real eigenvalue Az- of
T, and for all x G H we have
oo
2=1
Moreover, for every A ^ <r(T) and x G H, we have
oo / \
Proof: Let {e^} be the (countable) orthonormal basis of H from
Theorem 7.44. Note that, since {e2} is an orthonormal basis, for any x G H the
series ]P A2(#, e2)e; converges, because
m ~ m m
|5>(*>«.>.-|| = ?>•(*, c,-)|2 < imi^Kx.eOI2 - 0
2=n 2=n 2=n
as n, ?n —* oo. Also, if \\x\\ < 1, then for every n G N we have
|X)A,-(x)c,-)e,-|2 = ^AfK^eOl^llTH^Kx.eOI2
2=1 i=l 2 = 1
oo
< imia.x;i(*,e.-)ia = imi2-Nia.
2 = 1
OO
Thus, the operator G defined by G{x) = ]T A,-(a?,et-)ei G #(#) is conti-
2 = 1
nuous. From T(e2) = A^e; we obtain T(e2) = G(ez), so by linearity and
continuity, T — G on H.
Now consider some A G p(T). Since &(T) is closed, there is 6 > 0 such that
II m (# e-^ II2
dist(A, a(T)) > 8. Then |A2—A| > 6 for every i. Therefore ? v ' *; et- =
'U*=n A — Az- II
m \(x p-)|2 ™
? l\ ' 1 2 - 5 ? l(*>c0l2 due to the orthonormality of {et-}. Thus
i=n IA — A {I 2-=n
the series is convergent for every x G #, and we can define a linear operator
on H by G(x) = ? t^U- For ||x|| < 1, we have
A — Xi
if M, 
n \( M2 n
2 = 1
|A - A,-
8 = 1
< S-2?|(*)e,-)|2 = *-W<*-2;
2 = 1

224 7. Compact Operators on Banach Spaces
in particular, G is a bounded linear operator on H. For x — J2(x, ej)ej> we
have T(x) = J2\j(x, ej)ej and (XIh — T)(x) = J^(A — Xj)(x, e;)e;-; hence,
i i
using (e;,ej) = <52J-, we get
(XIH-T)(G(x)) = DA-Ai)(E^Tc«''ei)ei
Similarly, we show that G(\Ih — T) — Ih , and hence G = R(\).
D
Definition 7.47
Let H be a Hilbert space, and let T be an operator on H. T is called normal
ifTT* = T*T. T is called unitary if it is invertible and T~x = T*.
Clearly, every unitary operator is normal and every self-adjoint operator
is normal.
Proposition 7.48
Let H be a Hilbert space, T G B(H). T is normal if and only if \\T(x)\\ —
||jT*(#)|| for every x G //.
Proof: For x G #, we have
\\T{x)f-\\T\x)\\" = (rw.rwj-^wrw)
= (rr(x),*)-(Tr(x),x)
= ((T*T-TT*)(x),ar).
If T is normal, then the latter quantity is zero for every x G H, so ||T(a:)|| =
||r*(x)|| for every xeH.lf ((T*T-TT*)(x), x) = 0, then since T*T-TT*
is always self-adjoint, we have that T*T - TT* = 0 by Proposition 7.36
and thus T is normal.
?
Proposition 7.49
Let H be a Hilbert space. If T G B(H) is onto, then the following are
equivalent:
(i) T is unitary.
(ii) T is an isometry.
(Hi) (T(x), T(y)) ~ (x, y) for every x,yeH.
If the condition (iii) is satisfied for an operator T G B(H), we say that
T preserves the inner product.
Proof: By the polarization identities for a real or complex Hilbert space,
it follows that T preserves the inner product if and only if T is an

7. Compact Operators on Banach Spaces 225
isometry. Therefore, (ii) and (iii) are equivalent. If U is unitary, then
(T(x),T(y)) = (T*T(x),y) = (x,y) for every (a?,y), so T satisfies (iii).
If T satisfies (iii), then it is an isometry and onto, and hence T~l exists.
By (iii) also (T*T(x),y) = (x,y) for all x,y G H\ thus T*T = Ix and
T~l =T* follows.
?
Recall that if P is a bounded linear projection of a Banach space X onto
P(X), then X = P(X) 0 Ker(P).
Definition 7.50
Let H be a Hilbert space and P be a bounded linear projection of H onto
P(H). P is called an orthogonal projection ifKei(P) _L P(H).
Since we always have Ker(P) 0 P(H) = H and the two subspaces are
closed, we can equivalently write P(H) — Ker(P)-1 and Ker(P) = P(H)L.
Note that in Chapter 1 we proved that every closed subspace of a Hilbert
space H is complemented by an orthogonal projection.
Lemma 7.51
Let H be a Hilbert space and x,y ? H. If \\x + ay\\ > \\x\\ for all scalars a,
then (x, y) — 0.
Proof: We have ||z||2 < ||x + ay\\2 = (x, x) + \a\2(y, y) + 2Re(a(r/, x)).
Write a = rielt and (y,x) — r2e1^. Then |a|2(y, y) + 2 Re(a(y, x)) —
r\{y>y) + 2Re(rir2ez^+^). If we choose / such that t -f ^ = 7r, then
2Re(rir2e^t+^) = —2nr2 and thus we have —2rir2 + ^i||y|| > 0 for all
ri >0.
Then —2r2 -f rx > 0 for all r\ > 0, which is possible only if r2 = 0. Hence
D
Proposition 7.52
Let P be a bounded linear projection of a Hilbert space H onto P(H). Then
the following are equivalent:
(i) P is an orthogonal projection.
(ii) P is a self-adjoint operator.
(iii) P is a normal operator.
(iv) (x - P(x), P(x)) = 0 for all xeH.
(v)\\P\\ = l.
Proof: (i) => (ii): Follows easily from Ker(P) _L P(H) and x - P(x) G
Ker(P).
Clearly, (ii) => (iii).
(iii) => (i): By Proposition 7.48, we get Ker(P) = Ker(P*); also
Ker(P*) = P(H)L by Proposition 7.32, and hence Ker(P) = P(HI.

226 7. Compact Operators on Banach Spaces
(i) => (iv): If P(H) and Ker(P) are orthogonal complemented subspaces
of H, write x = y + z with y G P(#) and z G Ker(P). Then P(x) = y and
(x-P(*)>P(x)) = Br>y) = 0.
(iv) => (i): For y G P(#) and z G Ker(P), set x = y+z. Then P(x) = y,
# — P(x) — z, and we have (z, y) = [x — P(#)> P(#)) = 0; that is, y _L z.
Consequently, Ker(P) 1 P(H).
(i) => (v): Write zG# as z = y+z, where y G P(#), z G Ker(P).
Then y 1 z, so ||P(z)||2 = ||y||2 < ||y||2 + \\z\\2 = \\y + z\\\ Therefore
||P|| < 1. Since P(x) = x for x G P(X), we have ||P|| > 1. Thus ||P|| = 1.
(v) => (i): Write x = y + z with y G P(#), >? G Ker(P). Let a be an
arbitrary scalar. Then ||y|| = ||P(y + az)|| < ||y-faz||. By Lemma 7.51, we
have (y, z) = 0, and hence Ker(P) J_ P(H).
n
Theorem 7.53 (Spectral decomposition)
Let H be a complex infinite-dimensional separable Hilbert space. IfT is a
compact normal operator on H, then there exists an orthonormal basis {e;}
of Hj where each e2- is an eigenvector corresponding to an eigenvalue Xi of
T, such that for all x G H we have
i
In particular, if {Xn} denotes the set of all distinct eigenvalues ofT, and
Pn are the orthogonal projections of H onto the eigenspaces Ker(An/# — T),
then T = ]T}AnPn; where the sum converges in B{H).
n
PROOF: Consider the self-adjoint operator U = TT* = T*T. By
Theorem 7.44, there are (mutually orthogonal) eigenspaces Hn corresponding
to distinct eigenvalues Xn of U such that H = span((J Hn); in
particular, orthonormal bases of Hn (formed by eigenvectors of U) ordered into a
sequence form an orthonormal basis of H.
We claim that every Hn is an invariant subspace of T. Since UT —
TT*T = TTT* = TU, for h G Hn we get (XnIH - U)(T(h)) = T(XnIH -
U)(h) = T@) = 0; that is, T(h) G Ker(XnIH - 17) = Hn.
Each Hn has an orthonormal basis formed by eigenvectors of T.
Indeed, if An = 0, then U = 0 on Hn and also \\T(x)\\2 = (T(x)yT(x)) =
(T*T(x),x) = ([/(*),*) = 0; that is, T = 0 on #n. Thus, every nonzero
vector of iJn is an eigenvector. If An ^ 0, then Hn is a finite-dimensional
complex vector space and the existence of such a basis follows from linear
algebra and the fact that T\„ is normal.
Collecting the bases of all Hn) we get an orthogonal basis of H. The
convergence of X^'(x>e0e* ls proved as in Theorem 7.46, and a similar
proof shows that J2XnPn converges to T.
n
?

7. Compact Operators on Banach Spaces 227
Conversely, one can prove that if {e2} is an orthonormal basis of a
separable Hilbert space and A; —> 0, then the operator x ¦—> ^ A2(#, e2)ez- is a
compact normal operator.
Fixed Points
Definition 7.54
Let T be a map from a set K into K. A point xq G K is called a fixed point
ofT ifT(x0) = x0.
Let (Pjp) be a metric space and T: P —> P. T is called a contraction if
there exists g < 1 such that p(T(x), T(y)) < qp(x, y) for all xy y G P. T is
called nonexpansive if p(T(x),T(y)) < p(x,y) for all x,y € P.
Theorem 7.55 (Banach contraction principle)
Zef (P, p) be a complete metric space. IfT is a contraction from P into P,
then there exists a unique fixed point ofT.
Proof: Let a < 1 be such that p(T(x),T(y)) < qp(x,y) for x,y G P.
First, we show the uniqueness of a possible fixed point. If T(x\) = X\ and
T(x2) - x2 for xi,x2 G P, then p(zi,a;2) = p(T(xi),T(x2)) < qp(xux2),
which implies p{x\, x2) — 0 by q < 1. Therefore #i — #2-
To show the existence of a fixed point xq G P, we choose an arbitrary
point #i G P and define inductively #2 = T(a;i), xn+i = T(xn) = Tn(#i).
For n > 3, we have
p(xn+uxn) = p(T(xn),T(xn_1)) <qp(xnjxn_i)
< q2p(xn-1,xn„2) < < qn~1p(x2)Xi).
Therefore, for n > m G N, we have
p(Xn,Xm) < p(a?n,«n-l) + p(»n-l,»n-2) H h />(^m+l , xm)
< (qn~2 + qn'1 + • • • + gm-X*2, *i) < ?jp(*2, *i).
Consequently, {xn} is a Cauchy sequence, so there is x$ G X such that
#n —> xq. Since T is a continuous map, we have T(x0) = lim(T(arn)).
Because T(zn) = xn+i> we have #o = lim(a:n+i) = lim(T(?n)) = T(xo).
?
The following statement is a special case of a more general result
discussed, for example, in [DGZ3].
Theorem 7.56 (Browder, Gohde, Kirk; see [Kir])
Let K he a hounded closed convex subset of a separable Hilbert space H. If
T is a nonexpansive map from K into K, then there exists a fixed point for
T in K.

228 7. Compact Operators on Banach Spaces
Proof: Choose an arbitrary vq ? K. For s ? @,1), we define Ts(x) =
sT(x) + A - s)v0. Then TS(K) C K (K is convex) and \\T8(x) - T8(y)\\ <
s\\T(x) — T(y)\\ < s\\x — y\\. By Banach's contraction principle, for every
s ? @,1) there is a unique fixed point of Ts in K.
Let un be the fixed point of T1_jl in K. By the Eberlein-Smulian
theorem, K is a weakly sequentially compact set. Thus there is a subsequence
unj tending weakly to uq ? K in K. We will show that T(u0) — u§. First,
||iinj.-T(tio)||2 = IK- -u0 + u0-T(u0)\\2
= Hii^-tiolP + lltio-rCtio)!!2
+2Re(unj -u0,u0 -T(u0)).
Since unj -^ uq, we have
||tio - T(u0)\\2 = .Um (||«n. - T(Uo)||2 - \\unj - u0\\2). (*)
Since T is a nonexpansive mapping, we have
IK-tmh < IK-T(tin.)|| + ||r(tini)-r(ti0)||
< iK-r^oii + iK.-uoii,
and hence limsup(||wnj. -T(uq)\\ - \\unj - ti0||) < limsup \\unj -T(unj)\\.
3 3
Fromwnj ~T^l_^_){unj) - (l-?-)T(unj)+ ^-v0) we get T(unj)-unj =
~-(T{unj) — vo). Since {T(unj) — vo} C K — K, it is bounded in H, so
lim||T(uni) - tini|| = 0. Hence limsup(||wnj. - T(u0)\\ - \\unj - u0\\) = 0.
3 3
Using the boundedness of {un}, we obtain
limsup(||tini - T(u0)\\2 - \\unj - u0\\2)
3
= limsup(||wni - T(u0)|| - \\vnj ~ Ml) (\\unj - T{u0)\\ + \\unj - u0\\)
3
= 0,
so by (*) we have \\u0 — T(uo)\\ = 0 and u0 is a fixed point of T.
?
We will now show two results concerning the common fixed point of a
family of maps. Let K be a, set and let Q be a family of maps K —> K. Q
is called commuting if T1T2 — T2T1 for every Ti,T2 ? Q.
Theorem 7.57 (Markov, Kakutani)
Let K be a convex compact subset of a Banach space X. IfQ is a commuting
family of continuous affine maps from K to K, then there is xq ? K such
that T(xo) — xo for every T ? Q.
For Werner's proof [Wer] of this theorem based on the Hahn-Banach
theorem, we refer to Exercise 7.48.

7. Compact Operators on Banach Spaces 229
By a semigroup of maps on K we mean a family S of maps from K into
K such that Ik is in S and S1S2 G S whenever Si, 52 G S.
Theorem 7.58 (Ryll-Nardzewski; see, e.g., [Nam2])
Ze2 K be a weakly compact convex set in a Banach space X. Let S be
a semigroup of affine weakly continuous maps from K into K. If for all
x,y G K, x ^ y, we have 0 ? {s(x) — s(y)] s G S}, then S has a common
fixed point.
PROOF: We may assume that K is a minimal convex weakly compact set
invariant for S. Let M be a minimal weakly compact subset of K invariant
for S (Zorn's lemma). Then conv(M) = K. We must show that M is a
singleton. If not, then there are x, y G M, x ^ y. By our assumption, there
is e > 0 such that
IK*) - *(y)ll > e (*)
for all s G S. Since M is a weakly compact set in a Banach space, by
Theorem 8.28, there is a strongly exposed point u G M contained in a
relatively weakly open subset V of M with diam(V) < e. By the minimality
of M, we have M — {s{u)\ s G S} ; hence so(u) G V for some so G S.
Set z — \(x + y) G K. By the minimality of K, K — coiw{S(z) ).
By Theorem 3.41, u G S{z) . Hence, there is a net {sa} in S such that
$a{z) -* u- Since K is ^-compact, by taking subnets if needed, we may
assume that sa(x) —> a and sa(y) —> b. From sa(z) — |(sa(x) + Sa(y)),
we get w = I (a + 6) and a — b — u because t/ is an extreme point of K.
It follows that soSa(#) ~* 5o(a) = sq(u) G V. Since {sosa(zO} is a ne^
in M and V is relatively open in M, 50sa(^) G V eventually. Similarly,
{sos<x(y)} G V eventually, so, for some a, sqs^x) and s0sa(y) are both in
F. Then II
sosa{x) ~ sos(x{y)\\ ^ diam(V^) < ?, contradicting (*).
D
We will now quote two more theorems that are frequently used. However,
their proofs require an extensive preparatory work and are omitted; we refer
the reader to [Bol].
Theorem 7.59 (Brouwer; see, e.g., [Bol])
Let K be a compact convex subset of a finite-dimensional Banach space. If
/: K —> K is continuous, then f has a fixed point in K.
Theorem 7.60 (Schauder; see, e.g., [Bol])
Let A be a closed convex subset of a Banach space X. If f:A —> A is
continuous and K — f(A) is compact, then f has a fixed point in A.
Note that the assumptions are satisfied if K is a compact convex subset
of a Banach space and /: K —> K is a continuous map from K into K.
We show two applications of these theorems.

230 7. Compact Operators on Banach Spaces
Theorem 7.61 (Schauder)
Let H be a separable Hilbert space. IfT is a weakly sequentially continuous
map from Bh into Bh, then T has a fixed point in Bh-
Recall that T is weakly sequentially continuous if T(xn) —» T(x)
whenever xn —> x.
PROOF: Let {e2} be an orthonormal basis of H; put Hn — span{ei,..., en}
and Kn = Bh H Hn. Let Pn be the orthogonal projection of H onto Hn)
/ oo v n
namely Pn ( Y2 xiei) — J2 xiei- Then Tn — PnT maps the compact set Kn
\ = 1 ' 2 = 1
continuously into Kn. Therefore, by Brouwer's theorem, for every n there
is xn E Kn such that PnT(xn) = xn. Since 77 is reflexive, Bh is weakly
sequentially compact and thus there is a subsequence xnk that converges
weakly to some x E Bh. Since T is weakly sequentially continuous, we have
T(xnk) —¦> T(#). Because T(xn) = xn for every n, T(xnk) — xnk —* z, and
also T(znfc) ^ T(z). Thus T(a?) = x.
D
Theorem 7.62 (Aronszajn, Smith; see [ArSm])
Let X be an infinite-dimensional Banach space. IfT is a compact operator
on X, then T has a nontrivial invariant subspace.
Proof: Assume that ||T|| = 1 and put T = {S E B{X)\ ST = TS}. Note
that T is nonempty because Ix E T. Since T is a subspace of B(H), we get
span{5(y); S E T} = {?(?/); 5 E T} for any y E X. Note that if S E T,
then ST,TS E T. Thus, for any y E X \ {0}, 7 = {5(y); 5 E T} is an
invariant subspace of T and Y ^ {0}.
If there isyE^\{0} such that Y ^ X, we are done. Assume that no
such 2/ exists. Choose xq E X with ||T(aro)|| > 1- Since for any y E X \ {0}
we have Y = X, there is an operator S GT such that
||S(y)-*o||<l. (*)
Denote ffp = {x E X; ||z - a?0|| < 1} and note that 0 <? B0 U T(B0).
Because T(Bq) is a compact set not containing the origin, by (*) there
are operators Ti,...,Tn E T such that for every y E T(Bq) there is z
such that ||7i(y) — x0\\ < 1- For every y E T(Bq) and i < n, put A2 =
max{0,1 - \\Ti(y) - xp\\} and A(y) = ?A2(t/). Clearly, A(y) > 0 for every
yeT(B0). ForyETE0), put
*»> = 5».(»)
The map ^ = tp o T: Bo —+ Bo is continuous and tp(Bo) is compact, so
by Theorem 7.60 it has a fixed point on z0 / 0. Define a linear operator
by S0(z) = ? A7("o)/7;^). Then 50 E T and Z = Ker(/X - So o T)
A(T(*o)J

7. Compact Operators on Banach Spaces 231
contains z$. Since So and T commute, Z is easily seen to be a nontrivial
invariant subspace of T. Indeed, if z G Z', then z — SoT(z). Thus T(z) =
TSoT(z) = (S0T)(T(z)) and T(z) G Z. Because T|z has an inverse S0\z,
it is a compact isomorphism and thus dim(Z) < oo; in particular, Z ^ X.
?
Exercises
7.1 Show that if T, S are bounded linear operators on a Banach space X
and one of them is compact, then TS and ST are compact.
7.2 Let X be a Banach space and T G K{X)\T(X). Show that 0 G T(SX).
Hint: If the statement fails, then inf{||T(x)||;a: G S'x} > 0. Thus there is
C > 0 such that ||T(x)|| > C\\x\\; consequently, T(X) is closed in Y. By the
open mapping theorem, T(Bx) contains a neighborhood of zero in T(X)
that is then totally bounded. This is a contradiction with the fact that T
is not of finite rank.
7.3 Let Xy Y be Banach spaces, T G B(X, Y). Show that if T is continuous
from the weak topology of X into the norm topology of Y, then T is a
finite-rank operator.
Hint: Let W — {x; \fi(x)\ < 1} for some /i,...,/n G X* be such that
T(W) C Bx. Then T is 0 on f|/f HO)- Write x = fl/THO) © Z' where
dim(Z) < oo. Note that T(X) = T(Z).
7.4 Show that if T is a compact operator from X into Y, then T(X) is
separable.
Hint: T{Bx) is totally bounded, and hence separable.
7.5 Let X,Y be Banach spaces, T G B(X,Y). Show that T G /C(X,Y)
if and only if there is {<} C X* such that ||<|| -* 0 and ||T(z)|| <
supn l^nC^)! f°r every x e x.
Hint: The condition implies that T(BX) C C({x;} U {0}) is relatively
compact, using the Arzela-Ascoli theorem ({#n}U{0} is compact). Assume
that T is a compact operator. Then T* is compact and, by Exercise 1.60,
there is asequence {x*} C X* such that x* —> 0 and T*(BX*) C coUvfz* }.
Then, for all x G X,
||T(a:)||= sup /(T(x)) = sup (x) < sup |<(x)|.
/€BX* T*(BX*) n
7.6 Let X, Y be Banach spaces, T G B{X,Y). Show that AT G ?(X, Y) if
and only if there is A G Co and a bounded sequence {y* } in X* such that
||r(s)||<supn|A„||j6(*)|forallxeX.

232 7. Compact Operators on Banach Spaces
Hint: In the preceding exercise, write x* = ijf?]r||#n||-
7.7 (Grothendieck) Let X, Y be Banach spaces. Show that if T G JC(X, Y),
then there is a closed subspace Z of cq and R G IC(X, Z), S G IC(Z, Y) such
that T = SR (i.e., T factors compactly through a subspace of cq).
Hint: In the notation from the previous exercise, ||T(z)|| < supn |An| |?/*(ff)l
for all x G X. Put R(x) = (Xiy*(x)) G c0 for z G X. Define Z = R(X).
Then R G IC(X, Co) by the previous exercise (we use the coordinate func-
tionals). Define an operator S: R(X) —» Y by S(\iy*(x)) = T(#) for a: G X.
Clearly, 5 is continuous, so we can extend it to a bounded linear operator
Z-+Y. Then T = S o R, and since 5(J5Z) C T(BZ), S is compact.
7.8 Let X,Y be Banach spaces, X reflexive, and T G B{X)Y). Show that
if T is completely continuous, then it is compact.
Show that if T is not compact, then there is xn G X and e > 0 such that
zn -^ 0 and ||T(#n)|| > e for all n.
Hint: If T(arn) G T(Bx), let a:nfc —» a? G 5^ in X by the weak sequential
compactness of Bx- Then T(xnk) —> T(x), showing that every sequence in
T(Bx) has a convergent subsequence.
If T is not compact, by the first part get yn G X with yn —* y and
T(yn) 7^ TB/). Then use a?n = yn - y.
7.9 Let X, y be Banach spaces and T G $(X, Y). If Z is a closed subspace
of X, is T(Z) necessarily closed in Y1 Would it help if T were open?
Hint: No in both cases. Consider X = ?2 ® ?2, Y = ?2, Z = 0 0 ?2, and
the operator T from X onto y defined by T(x, y) — x -\- L(y), where L is
a compact operator with ||L|| = \. Then T(Z) is not closed in Y.
7.10 Let Xfc be the space Ck[0,1] endowed with the norm
11/11 = ?max{|/«)(*)|;*G [0,1]}.
Show that the identity map from X& into Xk-i is a compact operator.
Hint: The Arzela-Ascoli theorem.
7.11 Let T G B(X,?i). Show that if X is reflexive, then T is a compact
operator.
Hint: T(Bx) is weakly compact. In'^i, the weak and the norm compactness
coincide (Schur).
7.12 Let T G B(c0}X). Show that if X is reflexive, then T is a compact
operator.
Hint: Check T* using the previous exercise.

7. Compact Operators on Banach Spaces 233
7.13 Find a non-compact operator T G #(^2) such that T2 = 0.
Hint: Put T(e2n) — ?2n+i and T(e2n+i) = 0.
7.14 Let X be an infinite-dimensional Banach space. Show that there is a
bounded linear non-compact operator from X into cq.
Hint: Let fn G Sx* be such that fn —> 0 in X* (the Josefson-Nissenzweig
theorem). Define T(x) = (fi(x)). G Co for a: G X. T{Bx) is not compact
by Exercise 1.48.
7.15 Show that the space Co is isometric to a subspace of /C^)-
Hint: If (a2) G Co, then the operator T{x) — (a2#z) is in /C^)- Use the
proof of Proposition 1.28.
7.16 Show that /C(^) is not reflexive.
Hint: Use the previous exercise.
7.17 Show that /C(^) is not complemented in B(?2).
Hint: Assume there is a projection P from #(^2) onto ?(?2)- Using
Exercise 7.15 and the Sobczyk theorem, we obtain a projection Q from K{?2)
onto Co. Then the restriction of QP to the subspace of B(?2) of diagonal
operators, which is isomorphic to ^00, would yield a projection from ?oo
onto Co, a contradiction.
7.18 Show that K{?2) contains an isometric copy of ?2.
Hint: For y G l2, define Ty G B{?2) by Ty(x) = (f>mW Clearly,
I °° I
Ty G ?(^2), and y h-> Ty is a linear map. Using ^ a?t-j/t- < ||z||2||y||2i we
h=i I
get ||Ty|| < ||y||2- To get the opposite inequality, use x defined by X{ — yj.
7.19 Let X be a separable Banach space, and let {x{} be dense in Bx-
Define an operator T from X* into ?2 by T(f) = B7(xz)). Show that
T is a homeomorphism of (Bx*,w*) onto the compact set T(Bx*) in ^2
taken in its norm topology.
Hint: T is a compact operator. In norm compact sets in ?2, the pointwise
and norm topologies coincide.
7.20 Show that every compact metric space K is homeomorphic to a norm
compact subset of ?2 taken in its norm topology.
Hint: Put X — C(K) in the previous exercise. Note that C(K) is separable;
also xn —> x in K if and only if xn —* x in X*.
7.21 Show that Theorem 7.12 fails for A = 0.
Hint: T((#;)) = B~lXi) in ?2- It is compact; see Exercise 1.49.

234 7. Compact Operators on Banach Spaces
7.22 Let P be a bounded linear projection of a Banach space X onto
Y. Show that if Y is nontrivial, then o~(P) = {0,1}, both 0 and 1 are
eigenvalues, and for A / 0,1 we have (XIx — P)'1 — jlx + \(\-i)P-
7.23 Let T be a diagonal operator on l2 associated with a bounded
sequence of complex numbers {cz}; that is, T((#2)) = (c2-a?t-). Find the set of
all eigenvalues of T and the spectrum of T.
Hint: Show that A is an eigenvalue iff A is equal to one of the numbers
C{. Since cr(T) is closed, we have that {a} C (?(T). If A ^ {cz}, then
inf IA — C{\ > 0 and the diagonal operator with diagonal j^- (as an inverse
operator) shows that such A is not in or{T).
7.24 Let K be a compact set in the scalar field. Show that there is an
operator T E B(?2) such that <j(T) = K.
Hint: Consider a dense sequence {c,-} in K and the previous exercise.
7.25 Is there T G K(?2) with <r(T) = {1 + 1}~=1 U {0}?
Hint: Proposition 7.29.
7.26 (i) Let L be a left shift operator in ?2, L(xi,x2,...) = (a^, ?3,...).
Show that the set of all eigenvalues of L is the open unit disk.
(ii) Let R be a right shift operator on ?2) R(x\)X2J...) = @, x\, x2i.. •)•
Show that the set of all eigenvalues of R is empty.
(iii) Show that c(L) and (t(R) are both equal to the closed unit disk.
Hint: (i): The vector A,A,A2,...) is an eigenvector for A if |A| < 1. If
|A| > 1, then the equation for the eigenvalue is not solvable in ?2.
(ii): R is one-to-one and thus 0 is not an eigenvalue. If A ^ 0, then by
solving @, a?i, X2)...) = (A#i, A#2,...) we get X{ = 0 for all i.
(iii): Note that L = R*, so they both have the same spectrum up to
the conjugation. The norms of these operators are 1, so both spectra are
contained in the closed unit disk.
7.27 Consider the right shift R on ?2 and the diagonal operator D
associated with di = 2~\ Define a weighted shift operator T on the complex
space ?2 by T = R o D. Show that T is a compact operator with spectral
radius 0 and T is one-to-one. Thus, T has no eigenvalues and cr(T) — {0}.
Hint: Show first that D is a compact operator, and hence T is compact.
Write the explicit formula for Tn to see that ||Tn|| —> 0. Since the spectrum
must be nonempty, c(T) = {0}.
7.28 Find an operator T G B(?2) such that its spectrum a(T) consists
exactly of the points 0 and 1 and such that neither 0 nor 1 is an eigenvalue
forT.
Hint: Recall that a(S + filx) = &(S) + ft- Write i2 as the direct sum of
span{e22} and span{e2;+i}; then use the previous exercise.

7. Compact Operators on Banach Spaces 235
7.29 Let T be a Banach space and T G B(X). Prove that exp(T) - E ^r
exists, it is an invertible operator, and cr(exp(T)) = exp(<r(T)).
Hint: exp(T)exp(-T) = Ix. Show first that a(p(T)) = p{a{T)) for every
polynomial p.
7.30 Assume that T is a bounded linear operator from a Hilbert space H
into H such that (T(x),x) > (x,x) for every x ? H. Show that T is an
invertible operator on H.
Hint: By assumption, ||T(ar)||||a:|| > ||x||2; that is, ||T(x)|| > \\x\\ for
every x G H, so T is an isomorphism into (Exercise 1.27). We also have
(x,T*(x)) > (x,x), so ||T*(x)|| > \\x\\ and T* is an isomorphism into. By
an analog of Exercise 2.39, T is onto.
7.31 Let T be a linear operator on a Hilbert space H with (T(x),y) =
(x,T(?/)) for all x) y in H. Show that T is a bounded operator (the
Hellinger-Toplitz theorem).
Hint: Let xn -> 0 in H. Then |(xn,T(y))| < |M|||T(y)|| -> 0 for every
y E H. Thus, for every y e H, we have (T(a:n),y) = (xn,T(y)) -* 0.
This means that, by the Riesz representation theorem, the assumptions in
Exercise 3.45 are satisfied, and hence T is continuous.
7.32 It is not known whether an infinite-dimensional Banach space X exists
such that B(X) = span ({IX}UIC(X)). Show that a Hilbert space does not
satisfy this property.
It is also not known whether there is a Banach space X such that K(X)
is complemented in B(X). It is, however, known that there is a Banach
space X such that B(X) = span({I*} U S(X)) ([GoMa]), where S(X) is
the closed subspace of B(X) formed by strictly singular operators.
Hint: Let P be the orthogonal projection on the set of all even coordinates.
Assume that P = alx + K, where K is some compact operator. Then
K(x) = A — a)Ix(x) for every x that is supported on even coordinate
vectors. Since K is compact, we get a = 1, so P = Ix + K. Checking this
equality for vectors supported by the odd coordinate vectors, where P = 0,
we get that K — —Ix there, which is a contradiction with the compactness
of if.
7.33 Let T G B{H) be a self-adjoint isomorphism of a Hilbert space H
onto H. Show that if T is positive (i.e., (T(x), x) > 0 for all x E X), then
[x,y] — (T(x)}y) defines a new inner product on H and |||z|| = [a:, a;] 2 is
an equivalent norm on H.
7.34 Let H be a finite-dimensional Hilbert space. Follow the hint to show
that Ext(B&(h)) consist exactly of all unitary operators.
The statement is in fact true for all Hilbert spaces.

236 7. Compact Operators on Banach Spaces
In the hint, we use several facts from the operator theory. First, for every
T G B{H) there exists a polar decomposition T = US, where U is a unitary
operator and 5 is a positive operator; that is, (S(x)} x) > 0 for every x G H.
Second, if H is a finite-dimensional Hilbert space and S is a positive
operator on iJ, we can find an orthogonal basis of H so that the matrix of
S with respect to this basis is diagonal. Note that elements on the diagonal
have magnitudes at most ||5||.
Hint: Let T G Ext(B&(H)), and T — US be its polar decomposition. Then
\\S\\ < ||T||, \\U\\ = 1. We claim that S = /#. Indeed, otherwise we find
Si ^ S2 with ||5i||, ||52|| < 1 such that 5 = ^f^. Thus, T would not be
an extreme point.
On the other hand, let U be unitary. By the parallelogram equality,
\\{U + V)(x)f + ||(I7 - V)(x)\f - 2||G(z)||2 + 2||7(z)l|2.
Assuming \\U + V\\, \\U - V\\ < 1, we get V{x) = 0 for every x G H.
7.35 Let H be a separable Hilbert space. An operator T G B{H) is called
a Hilbert-Schmidt operator if there is an orthonormal basis {e2} of H such
that Ell^(e*)l|2 < °°- Show that if {/;} is another orthonormal basis of
/Mhen?||T(/0||2=EI|r(eOII2-
The number \\T\\Hs - (E II^C6*)!!2)* is called the Hilbert-Schmidt norm
ofT. Show that \\T\\Hs > \\T\\.
Hint:
Enroll2 = EEKT^)'ei)i2 = EEKe-T*(ei))i2
i i j i j
= EnT*(ei)ii2-
j
Thus
i i j j i
By Holder's inequality, we have
l|r(*)ll = ll?(*>*)r(*)ll < Z>,*)lim«)ll * NKEim^ll2)*-
7.36 Show that every Hilbert-Schmidt operator T on a Hilbert space H is
compact. Find a compact operator that is not a Hilbert-Schmidt operator.
Hint: Let'{et-} be an orthonormal basis of H; denote a^ — (T(ei), ej). Then
T(J2xiti) = T,(Eaijxi)ej> so \\t\\hs = (EE latf|2)*- Set aj = E^?e*-
j i i j i
Then aj G H for every j, \\T\\Hs = (EIKII2)^ and T(x) = EC*,^;.
j j
Clearly, T(Bh) C {x = E^*6** k*l ^ lla*ll f°r everY *'} and the set is
compact in H.

7. Compact Operators on Banach Spaces 237
For the second question, consider a diagonal operator on ?2 with diagonal
{ci} such that lim(cj) = 0 and ]T lc*|2 *s not convergent.
7.37 Let X be a Banach space, T G B{X) and x0 G X \ {0}. Put Y =
span{a?o, T(a?o), T2(aro), • • •}• Show that Y is an invariant subspace for T.
The vector xo is called cyc/zc for T if V = X. Show that T has no
nontrivial invariant subspace if and only if every xo G X \ {0} is a cyclic
vector for T.
7.38 Show that every operator on a nonseparable Hilbert space has a
nontrivial invariant subspace.
Hint: Use the previous exercise.
7.39 Let T be a contraction from a Banach space X into X. Show that
F — Ix — T is an open one-to-one map from X onto X.
Hint: Let yo G X be given. Consider the map ^(ar) = T(x) + yo. Note that
# G X is a solution of F(x) = yo iff 2? is a fixed point of the map (p. Observe
that ip is a contraction on X. Thus F is onto.
Next, show that B°(F(z), A - q)r) C F(B°(z, r)), where 5 is the
contraction constant of T and B°(z,r) is an open r ball centered at z: let
V ? 5?(.)(C1 ~ <?»• Put /(*) = x ~ F(*) + 2/ = V + r(x) for x G S°(*, r).
Then ||/(*) - z\\ = ||T(x) + y - z|| < ||T(x) - T(z)\\ + \\T(z) + y - z|| <
g||s-2:|| + ||y-F(z)|| < qr+(l-q)r = r. Therefore, f:B°{z,r) -> B°(z,r)
and / is a g-contraction. Thus there is w G B°(zi r) such that f(w) = iu.
This means w = w — F(w) + y; that is, i^iu) = y.
7.40 Define <p: R —> R by <?>(#) = x — arctan(x) + tt/2. Show that \<p(x) —
<p(y)\ <\x — y| for #, y G R and there is no fixed point for cp on R.
Hint: The mean value theorem.
7.41 Find an example of a non-complete metric space and a contraction
on it without a fixed point.
Hint: Think of open interval @,1) and linear functions.
7.42 (Alspach) Consider K = {/ G ?i[0,1]; 0 < / < 2, ? f dt = 1} and
T\K-*K defined as
Tif\U\-[ minB/B*),2) . ifO<*<^,
WW - I max@) 2/B* - 1) - 2) if \ < t < 1.
Show that T is an isometry (in particular, a nonexpansive map) on the
weakly compact set K without any fixed point.
It is an open problem whether a nonexpansive map on the unit ball of a
(super)reflexive Banach space has a fixed point.
Hint: The weak compactness of K follows from the following criterion: A
set M C L\ is weakly relatively compact if for every e > 0 there is S > 0

238 7. Compact Operators on Banach Spaces
such that for every A C [0,1] satisfying A (A) < 6 and every / G M we have
fA |/| dX < e (where A is the Lebesgue measure; see, e.g., [DuSc]).
Assume by contradiction that / G K is a fixed point of T. Prove by
induction that A({*; 2~k < f(t) < 2~k+1}) = 0 for k = 0,1,.... Thus
A({t; /(*) = 0}) = A({2; f(t) = 2}) = \. Next, show that A({t; /(*) = 0}D
[^r, ^]) = ^ for arbitrary tEN,0<n<2fc-l. This is a contradiction
with the Lebesgue density point theorem for the set {t\ f(t) — 0}.
7.43 Let C be a closed convex bounded subset of a Banach space X. Show
that if T:C —* C is a nonexpansive map, then inf{||ar —T(x)||; x G C} = 0.
Hint: Pick z G C, ? > 0, and consider T?{x) = ez + (I - e)T(x). Then
T?:C —» C since C is convex. Moreover, T^ is a contraction. Therefore
there is x? G C such that xe — T?(x?). Then we have ||se — T(x?)|| =
||ez + A - e)T(a:e) - T(xe)|| = e\\z - T(a?e)|| < e diam(C).
7.44 Use the previous exercise to show that a nonexpansive map on a
convex compact set in a Banach space has a fixed point.
Hint: The infimum is attained.
7.45 Define a map T\ SCo —> SCo by T{x\^ #2> • • •) — A, a?i, a?2 • • -)- Show
that T is an isometry without a fixed point.
7.46 Let A = {x(t) G C[0,1]; 0 = x@) < x(t) < x(l) = 1}. Define a map
T: A —> A by T(x)(t) = tx(t). Show that A is a closed bounded convex set
in C[0,1] and T is a nonexpansive map (in fact, ||T(ar) — T(y)\\ < \\x — y\\
for every x ^ y). Yet, T has no fixed point.
Hint: Solve tx(t) = x(t) for t G [0,1).
7.47 Define Ur: L2(R) -> L2(R) by Ur(f):t ^ /(t-r). Show that {Ur\ r G
R} is a commutative group of unitary operators.
7.48 (Markov-Kakutani) Let K be a compact convex set in a locally convex
space E. Show that if T: K —> K is a continuous affine map, then T has a
fixed point.
Hint: (Werner [Wer]): By contradiction, assume that the intersection of the
diagonal A = {(x} x)\ x G K] of K x K with the graph T of T is empty.
By the Hahn-Banach theorem, there exist continuous linear functionals l\
and ?2 on E and numbers a < {3 such that
h(x) + h(x) <a<C< h(y) + h(T(y))
for all x, y G K. Consequently, I2 (T(x)) — hix) >/? — a for all a: G K.
Iterating this inequality yields I2 (Tn(x)) — h^x) > n(/?— a) —»• 00 for arbitrary
a? G if. Thus, the sequence {l2(Tn(x))} is unbounded. This contradicts the
compactness of ^(/i).
The full statement of the Markov-Kakutani theorem follows from this
by considering the sets of fixed points of mappings T{, i G /. Indeed, if Ki

7. Compact Operators on Banach Spaces 239
is the set of all fixed points of T{, we have Ki ^ 0 and K{ is compact and
convex. From the commutativity we get Ti(Kj) C Kj. Hence Ti\ has
a fixed point and Ki D Kj ^ 0. Thus f| K{ ^ 0 for every finite F C I.
Therefore p| Ki ^ 0, which proves the result.
*€/
7.49 Show that there is a continuous map from Bt2 into Bi2 that has no
fixed point. This means that the requirement of finite dimension cannot in
general be dropped in Brouwer's theorem (see [Dugl]).
Hint: (Kakutani) Define y>{xi,x2,...) = (|A — IMh), xi,x2, • • •)•
7.50 A retract of A onto B C A is a map from A onto B whose restriction
to B is the identity map on B.
Show that the following two statements are equivalent for every n:
(i) There exists a continuous retract of Bi* onto Sg».
(ii) There is a continuous map of B^ into B^ without a fixed point.
By Brouwer's theorem, both statements fail. The situation is different
if X is an infinite-dimensional Banach space. Then Sx is even a Lipschitz
retract of Bx (see, e.g., [BeLi]).
Hint: If r is such a retract, consider the map x >—> —r(x) and note that
any fixed point of it would be necessarily on Si*. If / is a map of Bi* into
Bi* without a fixed point, consider the map that maps x E Bi* onto the
intersection of the ray from x to f(x) with ?g».
7.51 Let f(x:y) be a continuous bounded function on an open set Q in
R2 that is uniformly Lipschitz in y on fi (there is a constant K such that
l/(z,2/i) -f(x,y2)\ <K\yi - y2\ for every (x, yi), (x, y2) G ?2).
Let (#0) yo) 6 fi. Show that there is 8 > 0 such that on [xq — 8, xq + 8]
there is a unique continuously differentiable solution to -^ — f(x,y) with
the initial condition y(xo) — yo-
Hint: Observe that y[x) is a solution if and only if y{x) — t/o +
Ixo -f (*> 2/@) dt for x ^ fro ~ ^ xo + 6].
There is M > 0 such that |/(z,y)| < M for every (z,.y) ? ?2. Choose
<5 > 0 such that K8 < 1, set i" = [xq — 5, a?o + ?], and
# = {(*,!/) ?r2; \x-x0\ < 8,\y-yQ\ < M8} C Q.
Consider the closed subset Y of C(I) formed by all functions y{x) for which
y(xo) = yo and \y(x) — yo\ < M8. Then Y is a complete metric space in
the metric induced from C(I). Finally, define a map T on Y by
T(y):x^yQ+ f f(t,y{t))di
Jxq
for y{x) E Y. Check that indeed T maps Y into Y.

240 7. Compact Operators on Banach Spaces
Using the Lipschitz property of /, we get that T is a contraction, and
hence it has a fixed point. But T(y)(x) = y(x) is equivalent to y being a
solution on I.
7.52 (the Peano theorem) Let f(x,y) be a continuous function on R =
{(x,y) G R2; \x — xq\ < a,\y — yo\ < b}. Let M — max |/(z,y)| and
(x,y)eR
8 — min{a, -^}. Show that on the interval [xo — 8) xq + 8] there exists at
least one solution to the equation -J^ — f{x, y) that satisfies y(xo) = yo.
Hint: A function y is a solution if and only if y(x) = yo + fx f(t, y(t)) dt.
Set I = [xq — 8, xq + 8] and let B be the closed ball in C(I) centered at the
constant function yo with radius b. Define a map T on B by
T(y)=y0+ [ f(t,y(tj)dt.
Jxq
T is continuous: If yn(x) converges to y(x) in jB, then by the uniform
continuity of /(#, y) on the rectangle R we have that f{x) yn(x)) —* /(#, y(x))
uniformly on I. Therefore, T(yn)(x) —> T(y)(x) uniformly on I.
The total boundedness oiT(B) is proved directly using the Arzela-Ascoli
theorem. Thus T(B) is compact and, by Schauder's theorem, T has a fixed
point y(t) G B.
7.53 (Korovkin) Let {Tn} be a sequence of bounded linear operators on
C[0,1] that are positive; that is, Tn(/) > 0 if / > 0 on [0,1]. Assume that
T„(l) -> 1, Tn(x) -+ x, and Tn(x2) -+ x2 in C[0,1]. Show that Tn(f) -+ /
in C[0,1] for every / G C[0,1]. Is it true that \\Tn - IC[oti]\\ -> 0?
Hint: To prove Tn(/) —> /, it is enough to find for every t G [0,1] and
e > 0 some no G N and 8 > 0 satisfying |Tn(/)(r) — /(r)| < e whenever
\r — t\ < 8, n > n0.
Since / is continuous, there are quadratic functions (pi,<p2 G C[0,1]
satisfying (f2 < f < <Pi on [0,1] and \<pi(r) - <p2(j)\ < e for \r - t\ < 8.
Thus, Tn(pi - /)(r) > 0, Tn(f - <p2)(r) > 0, and Tn(y>i)(r) - ^(r),
^(v^X?") —* ^(t) as n —> 00 for |r — t| < 6. Thus, for n large enough,
<Pi(r) > Tn(f)(r) > <p2{r) on [0,1], so |T„(/)(r) - /(r)| < e.
\\Tn — ic[o,i]|| —*¦ 0 may fail; consider Tn: f »-» f(n\ where /(n) is the
broken line agreeing with / at its nodes at points ~, 0 < i < n.

8
Differentiability of Norms
Definition 8.1
Let f be a real-valued function on an open subset U of a Banach space X.
Let x E U. We say that f is Gateaux differentiable at x if there is F E X*
such that
limf(^th)-f(x) =
t-*o t v
for every h E X.
We say that f is Frechet diiferentiable at x if this limit is uniform for
hesx-
We call F the Gateaux or the Frechet derivative (or differential) of f at x
and denote it by F = f(x).
We say that f is Gateaux (resp. Frechet) differentiable (or smooth) on U
if f is Gateaux (resp. Frechet) differentiable at all points of U.
We say that f is Cfl-smooth on U if it is Frechet differentiable at every
point of U and the map x i—> f(x) is continuous as a map U —> X*.
Given x E X and h E Sx, if the limit
lim/(»+^)-/(»)
t—o t
exists and is finite, we call it the directional derivative of / at x in the
direction h.
A norm || • || on a Banach space X is called Frechet (resp. Gateaux)
differentiable if || • || is Frechet (resp. Gateaux) differentiable at every point
of X \ {0}. The reason for this is that a norm is never differentiable at 0.

242 8. Differentiability of Norms
Since differentiability conditions for a norm are homogeneous, a norm
is differentiate at x if it is differentiate at Xx for some scalar A.
Consequently, it is enough to check the differentiability at points of Sx-
Consider a norm || • || of a Banach space X. If it is Gateaux differentiate
at x e Sx, then from | Hg+**IHgll | < Mil = i for ft G Sx we get that
F = \\x\\f (the derivative of || • || at x) satisfies \\F\\ < 1. On the other hand,
considering ft = x, we get F(x) = 1. Thus, F is a supporting functional of
Bx at x.
Definition 8.2
Let U be a convex subset of a vector space V. We say that a function
/: U -* R is convex if f(Xx + A - X)y) < Xf(x) + A - X)f(y) for all
x,yeU and Xe [0,1].
For instance, every norm of a normed space X is a convex function on
X. Also, every linear functional on X is a convex function. More generally,
every positively homogeneous sublinear functional is convex.
Note that a function f:U —> R is convex if and only if the function
f >_> fix+tyj-fy*) is increasing in t for all x G U, y G V, and all t for which
x + tyeU.
Now assume that / is a convex function on an open convex subset U of
a Banach space X. It follows that for all x G U and ft G Sx, the one-sided
limits F+(h) = lim H°+th)-H*) and F~(h) = lim /(*+**)-/(«) exist
7 t—o+ * t—o- *
finite. Moreover, we have F+(h) > F~(h) and it is easy to check that F+
is a subadditive function in ft, F~(h) is a superadditive function in X, and
both are positively homogeneous.
Lemma 8.3
Let f be a convex function defined on an open convex subset U of a Banach
space X that is continuous at x G U. Then f is Frechet differentiable at x
if and only if
lim f(x+th) + f(x-th)-2f{x) = Q
t—o t
uniformly for ft G Sx -
An analogous result is true for the Gateaux differentiability. Note that
by the triangle inequality and convexity of / we always have f(x + th) +
f(x-th)-2f(x) >0.
Proof: If / is Frechet differentiate, then F+(h) ~ F~(h) for all ft, and
the claim follows from the following equality for t > 0:
f(x + th) - f(x) f(x - th) - f(x) = f(x+th) + f(x-th)-2f(x)
t -t ~ t
On the other hand, if the limit is 0, by this equation we have F+ —
F~. Then F = F+ = F~ is a linear functional. We must show that it is

8. Differentiability of Norms 243
bounded. Since / is continuous at x, there are <5, K > 0 such that / < K
on z + 6BX- By convexity, /(«+**)-/(«) < fLzl^l for an |/| < 6j h e Sx.
Thus F(h) < ^f^ for all h G SX-
D
Note that for a convex function /, the function t h-> f(^+th)+f{x-th)-2f{x)
is increasing in t for t > 0.
The uniformity in h G Sx in the case of Frechet differentiability allows
us to write the definition as lim \\y\\~ = 0> and similarly for the
limit in Lemma 8.3.
Lemma 8.4 (Smulian lemma [Smu])
Let (X, || • ||) be a Banach space, and let x ? Sx-
(z) || • || is Frechet differentiable at x if and only if lim ||/n — gn\\ = 0
n—+oo
whenever fn,gn G Sx* satisfy lim (fn(x)) = lim (gn(x)) = 1 if and only
n—>oov y n-*oov '
if {fn} C Sx* is convergent whenever lim (fn(x)) — 1.
n—foo
(n) || • || is Gateaux differentiate at x if and only if (fn — gn) —* 0 in X*
whenever fnjgn G Sx* satisfy lim (fn(x)) = lim (^n(x)) — 1 if and only
n—KX) 7 n—+oo
if there is a unique f G Sx* sz/cA Matf /(a:) = 1.
Proof: (i): Let || • || be Frechet differentiable at xGSj. Given e > 0, by
Lemma 8.3 there is 8 > 0 such that ||ar + A|| + \\x — h\\ < 2 + e\\h\\ whenever
INI < *•
Let fn,gn G Sx* be such that lim(/n(?)) = \im(gn(x)) = 1. Choose
no G N such that for n > no we have max{|/n(z) — 1|, \gn(x) — 1|} < eS.
Then, for n > no and \\h\\ < 6, we have
(fn - 9n)(h) = fn(x + h) + gn(x - n) - fn(x) - gn{x)
< \\x + h\\ + \\x - h\\ - fn(x) - gn(x)
< 2-fn(x)-gn(x) + e\\h\\
< |1 ~ fn(x)\ + |1 - gn(x)\ + e\\h\\ < 3e6.
Hence, for n > no, we have
\\fn-gn\\ = sup(/n-^n)(A)= sup (/n"f)(<5/l)<3g
\\h\\ = l \\h\\ = l 0
and lim ||/n - gn\\ = 0.
n—»-oo
Assume that || • || is not Frechet differentiable at x G Sx • By Lemma 8.3,
there is e > 0 and a sequence {hn} with lim||nn|| = 0 such that for every
n G N, ||ar + hn\\ + \\x - hn\\ > 2 + e\\hn\\.
For every n, choose fn,gn G -Sx* such that fn(x + /*n) — \\x + hn\\ and
9n(x-hn) = ||a?-An||. Note that |/n(/*n)| < \\hn\\ and |||ar + hn\\ - \\x\\\ <

244 8. Differentiability of Norms
\\hn\\ for every n, so fn(hn) —> 0 and \\x -f hn\\ —* 1. Hence
lim(/n(^)) = lim(/n(a + K) - fn(hn)) = hm(\\x + hn\\ - fn(hn)) = 1.
Similarly, we get \im(gn(x)) = 1. We also have
(fn - 9n)(hn) = fn(x + hn) + gn(x - hn) - (fn + gn)(x)
< \\x + hn\\ + \\x-hn\\-2>e\\hn\\.
Thus ||/n - gn\\ > e for every n.
The third statement is a reformulation of the second one.
The first equivalence of (ii) follows similarly. Assume the uniqueness of
the supporting functional at a?. If there are fn,gn G Sx*, V G Sx, and
e > 0 such that fn(x) -» 1, 0n(z) —> 1, and |(/n - #n)(y)| > ?, we take
tu*-cluster points /, # of {/n} and {gn} (by the w*-compactness of Bx*).
Then /(a?) = ^r(ar) = 1 and |(/ — g)(y)\ > e. Thus there are two support
functional for Sx at x, a contradiction. The other implication is obvious.
?
Corollary 8.5
If the norm || • || of a Banach space X is Frechet differentiable, then it is
C1-smooth on X\{0}.
Proof: Let x,xn G X \ {0}, xn —* x. Denote by /n, resp. /, the derivative
of || • || at xn, resp. x. Then /, fn G Sx, fn(xn) = ||a?n||, and f(x) = \\x\\.
Set yn = ikii ' ^= m ; then fn(yn)= lj f(y)= L
We easily check that fn(y) -^ 1, so by Lemma 8.4 we get fn —» /.
D
Theorem 8.6
Let X be a Banach space. If the dual norm of X* is Frechet differentiable,
then X is reflexive.
Proof: By Corollary 3.56, it is enough to show that every / G X* attains
its norm on Bx- Given / G Sx*, choose xn G Sx such that f(xn) —> 1. By
Lemma 8.4, {xn} C X** is convergent to some x E Sx- Clearly, f(x) = 1.
?
Theorem 8.7 (Kadec [Kad2], Restrepo [Res])
Let X be a separable Banach space. If X admits an equivalent Frechet
differentiable norm, then X* is separable.
Proof: The set B = {\\x\\f] x G Sx} is a James boundary of X. Since Sx is
separable, by Corollary 8.5, B is also separable, and hence X* is separable
by Corollary 3.49.
?

8. Differentiability of Norms 245
An alternative proof is to observe that C — {\\x\\'\ x ^ 0} is separable
and contains all norm-attaining functionals; hence C is dense in X* by the
Bishop-Phelps theorem.
Lemma 8.8
Let (X, || • ||) be a Banach space, and let ||| • ||| be an equivalent norm on X*.
I • I is a dual norm to some equivalent norm on X if and only if ||| • ||| is
w*-lower semicontinuous.
Recall that a norm | • |* on X* is a dual norm to some norm | • | on X if
for every / G X* we have |/|* = sup{/(#); \x\ < 1}.
Proof: As a supremum of i/Acontinuous functions x: f »—> f(x) for x G Sx,
every dual norm is w*-lower semicontinuous.
Assume that ||| • ||| is a w*-lower semicontinuous equivalent norm on X*
and denote by B its closed unit ball. Then B is u>*-closed and, by the
bipolar theorem, B = (B0)°. It follows that ||| • ||| is the dual norm to the
(equivalent) norm given by the Minkowski functional of Bo.
?
Theorem 8.9 (Leach, Whitfield [LeWh])
Let X be a separable Banach space. If X* is not separable, then X admits
an equivalent norm that is nowhere Frechet differentiable.
Proof: Since Bx* is not norm separable, there is n such that Bx* contains
no countable --net. Thus, a maximal --separated set S obtained by Zorn's
lemma is uncountable. On the other hand, Bx* in the u>*-topology is a
metrizable compact and thus a separable space. Therefore, (Bx*,w*) is
second countable.
Let A — {Ak} be a countable base for (Bx*>w*) (i.e., every w*-open
set is a union of a sub collection of A). Collect all Ak such that S 0 Aj~
is countable. Denote the collection of these Ak by B. Let S be the set of
all elements of S that lie in some of Ak, Ak G B. Clearly, S is countable,
and if s G S\S, then every ^-neighborhood of s contains an uncountable
number of elements of 5.
LetC^conVi;*[E\5)u(--E\5))] and U* = Bx* +C. Both Bx* and
C are iu*-compact, so U* is a w*-compact convex and symmetric set. Its
Minkowski functional is thus w*-lower semicontinuous; hence U* is the dual
ball of some equivalent norm || • || on X by Lemma 8.8. We will show that
|| • || is nowhere Frechet differentiable. By Smulian's lemma, it is enough to
show that every set of the form U* H {/; f(x) > \\x\\ — S}, where x G X
and 6 > 0 are arbitrary, has diameter greater than or equal to -.
Let D be such a set. Then D contains an element of the form 6 + c, where
6 G Bx* and c = ? Afc2- with A,- > 0, ? At- - 1, and a G (S\S)u(-(S\S)).
Thus 6 -f c — J2 ^i(° + ci)- Since {/; f{x) > ||z|| — 8} C X* contains a
convex combination of points 6 + Cj, it must contain at least one of them.
Therefore, there is i such that b+Ci is in D. Since {/; f(x) > \\x\\ — 8 — f(b)}

246 8. Differentiability of Norms
is a tu*-neighborhood of ct-, it also contains d E Bx* with \\d — ct-|| > ^.
Then ||F + cz) — F + d)\\ > ^ and (b + d) E D, which completes the proof.
D
Definition 8.10
A norm || • || of a Banach space X is called strictly convex (or rotund) if
Ext(Bx) = Sx.
Fact 8.11
Let (X, || • ||) be a normed space. The following are equivalent:
(i) || • || is strictly convex.
(ii) Ifx,yE Sx satisfy \\x + y\\ = 2, then x = y.
(Hi) Ifx.yeX satisfy 2\\x\\2 + 2\\y\\2 - ||z + y||2 = 0, then x = y.
(iv) If x,y ^ 0 satisfy \\x + y\\ = ||a?|| + ||t/||, tfAen x — \y for some A > 0.
Proof: (i) <=> (ii) is easy; \\x + y\\=2 means that \(x + y) E Sx-
(ii) ==> (iii): Note that
2|H|2 + 2|M|2-||x + 2/||2 > 2|H|2 + 2|b||2-(|M| + ||y||J
= (INI-lbliJ>o.
Thus, if 2||x||2 + 2\\y\\2 - \\x + y\\2 = 0, then ||x|| = ||y||. Hence we may
assume that x, y E Sx] we get \\x -f y\\ — 2 and (ii) implies x — y.
(iii) =>• (ii) is immediate.
(iv) =» (ii): If a?,y E Sx and ||* + y\\ = 2, then ||x|| + ||y|| = \\x + y||
and hence x — y.
(ii) =$> (iv): Let ||? -f y|| = ||a:|| + \\y\\ for some x, y / 0. We may assume
that 0 < ||z|| < \\y\\. Then
2 > ||»/I|ar|| + S//IMI|| > ||*/ll*IH-»/||*|||| — ||Z//II«|| — Z//llz#ll||
= (l/|W|)||2r + y||-||y||(l/||x||-l/||y||)
Thus ||*/||x|| + y/||y|||| = 2, and then x/||*|| = y/||y||.
a
Fact 8.12
Let (X, || • ||) be a normed space, and let || • ||* be the dual norm of X*. If
|| • ||* is strictly convex, then || • || is Gateaux differentiate.
Proof: By Lemma 8.4 (ii), we need only show that for x E Sx there is
a unique / E Sx* such that f(x) = 1. Let x E Sx and /, g E Sx* be
such that f(x) = ^(x) = 1. Then 2 > ||/ + g\\* > (f + $r)(x) = 25 that is>
||/ -f g\\* = 2, and from the strict convexity of || • ||* we obtain / = g.
D

8. Differentiability of Norms 247
Theorem 8.13
Every separable Banach space admits an equivalent Gateaux differentiable
norm.
Proof: Let X be a separable Banach space. Let {xn} C Sx be dense in
Sx- Define a norm ||| • ||| on X* by
I/in2 = \\f\\2+J22~if2^)'
where || • || is the original dual norm of X*. Using Lemma 8.8, we find that
I • I is the dual norm to an equivalent norm || • ||x on X. It is enough to
show that If • I is strictly convex.
Let /, g E X* and 2|||/||2 + 2|01||2 - |||/ + g|||2 = 0. Since 2||/||2 + 2||ff||2 -
11/ + 9\\2 > 0 and 2f2{xi) + 2g2(xi) - (/ + gf{xi) > 0 for every i E N, we
get 0 = 2/2(a;i) + 2g2(xi) - (/ + gJ{xi) = (/ - gf{Xi). Thus / = g on a
dense set, so / — g and ||| • ||| is strictly convex. Then use Fact 8.12.
?
Theorem 8.14 (Mazur; see, e.g., [Phe2])
Let X be a separable Banach space. If f is a continuous convex function on
X, then the set G of all points of Gateaux differentiability of f is a dense
G§ set in X.
Proof: We will prove the theorem for / being an equivalent norm || • || on
X. Let A be the set of all points in X where the norm || • || of X is not
Gateaux differentiable. We will first show that A is an Fa set.
Let {xn} be dense in Sx- For n, m G N, let
Am^n = xEl\ {0}; there are /, g in Sx* supporting Bx at #/||tf||,
(f-g)(xn)>-Y
m j
By Smulian's lemma, A = [j Am^n U {0}. We claim that Am)n is closed
772,n
for every m,n. Fix ra,n G N. Let z^ G Am^n and lim(^) ~ z m X. Let
fk,9k G Sx* be such that fk(zk) = gk{*k) - \\zk\\ and (fk - gk)(xn) > ^.
Because Bx* is a u;*-metrizable compact, we may assume that fk—*f and
gk ^ g in X*. Then /, g G Bx* and
\f(z)-fk(zk)\ < \(f-fk)(z)\ + \fk(zk-z)\ < |(/-/jb)(z)| + ||zt-z|| - 0.
Moreover, lim(fk(zk)) = lim||z;b|| = ||z||. Thus f(z) = \\z\\ and / G Sx*-
Similarly, g G Sx* and g(z) - ||z||. Also, (f-g)(xn) - lim (fk-gk)(xn) >
K-+00
— . Therefore z G im)n, which shows that Am^n is closed. We now show
that X \ Am n is dense in X. Fix m, n G N.
Let xq ? X and e > 0 be given. Consider the real-valued convex function
F defined for t G R by F(t) = ||a?o +^n||- F is differentiable at all but

248 8. Differentiability of Norms
countably many points t G R (see Exercise 8.5). Let |?'| < e be such that
F is differentiate at xq + t'xn. We claim that xo + t'xn ? Am)n. Indeed,
if xo + t'xn G Am^n) then there are /,g G Sx* such that f(xo -f /'?n) =
||x0 + *;*„|| - F(t'), g(x0 + t'xn) = F(t') and (/ - ^)(xn) > ?. Then
the restrictions of / and # to the line {xo -f ^n} would provide for two
different tangent lines to the graph of the function F at xo + t'xn. This
is a contradiction with the differentiability of F at this point. Therefore
xo + t'xn ? -Am>n, and X \ Am^n is dense in X.
Thus every X \ Am}Tl is open and dense in X, so
X \ A = X \ (U Am,n U {0}) - f](X \ Amtn) H (X \ {0})
is a dense G$ set in X by the Baire category theorem.
a
Definition 8.15
We say that a Banach space X is weak Asplund if for every convex
continuous function f on X, the set of all points of Gateaux differentiability of
f contains a dense Gs set in X.
Theorem 8.14 gives that every separable Banach space is a weak Asplund
space.
One can define a convex continuous function / on a nonseparable Hilbert
space H such that the set D of all points of Gateaux differentiability of /
is not a Gs set ([HSZ]). On the other hand, D contains a dense Gs set since
H is a weak Asplund space.
We now proceed by proving results analogous to Theorems 8.13 and 8.14
for Frechet differentiability.
Definition 8.16
A norm || • || of a Banach space X is called locally uniformly rotund (LUR)
if for all x,xn ? X satisfying limB\\x\\2 +2\\xn\\2— \\x-\-xn\\2) =0 we have
lim \\xn — x\\ — 0.
n—>-oo
Note that a norm || • || on X is LUR if and only if lim||a?n — x\\ — 0
whenever xn)x G Sx are such that lim||arn + x\\ — 2 (Exercise 8.43). It
is also obvious that LUR implies strict convexity (rotundity), but it is
a strictly stronger notion (Exercise 8.51). Note that it follows from the
parallelogram law that the norm of a Hilbert space is LUR.
Theorem 8.17 (Kadec; see, e.g., [DGZ3])
Every separable Banach space admits an equivalent LUR norm.
Proof: Let X be a separable Banach space. Let {xn} C Sx be dense in
Sx, and let {/n} C Sx* be a separating family for X. For n G N, put
Fn = span{#i,..., xn} and note that dist(#, Fn) —> 0 for all x ? X. Define

8. Differentiability of Norms 249
a norm ||| • ||| on X by
oo oo
IWI2 = Ikll2 + ?2_n dist(*' F»? + ? 2_n/?(*),
where || • || is the original norm of X and the distance is in the norm || • ||.
Since dist(#, Fi) is a positive homogeneous subadditive function, ||| • ||| is an
equivalent norm on X.
We will show that ||| • ||| is LUR. To this end, assume that X]Z) x G X are
such that lim B|H2 + 2|||zjk|||2 - \\x + xkf) = 0.
k —>-oo
Since the expression in this limit is nonnegative for all the functions
involved in the definition of ||| • |||, it implies the existence of the following
limits:
lim B||x||2 + 2||zfc||2 -\\X + Xk\\i)=0,
k —»-oo
lim Bdist(ar, Fnf + 2dist(xk,FnJ - dist(x + xk, Fnf) = 0 for every n,
k—*oo
lim Bfl{x) + 2f2(xk) - ft(x + xk)) = 0 for every n.
k—^-oo
As in the proof of (ii) => (iii) in Fact 8.11, we conclude that then
lim \\xk\\ = ||z||, A)
k—>oo
lim (dist(a?jfc, Fn)) — dist(#, Fn) for every n, B)
k—>oo
lim (fn(xk)) - fn(x) for every n. C)
k—+oo
Since {fn} is separating, the topology of pointwise convergence on {/n}
is a Hausdorff topology in X. We will show later that {xk} U {x} is norm
compact. Therefore, on {xk} U {x} the topology of pointwise convergence
on {/n} is equivalent to the norm topology. Thus, C) implies that lim||a?fe —
#|| = 0 and the proof is complete. It remains to show that {xk} U {x} is
norm compact.
Using A), choose K > 0 such that \\xk\\ < K for every k. Let e G @,1)
be given. Choose n G N such that dist(#,Fn) < ?, and choose a finite
?-net F in (K + l)#Fn- Using B), choose fco such that dist(xfc, Fn) < e for
A: > &o- We claim that {#i, #2,..., a?fc0} U jP is a 2^-net for {#&}. Indeed,
for every fc > k0 there is arj. G Fn such that ||#fc — x'k\\ < e. Since \\xk\\ < K
for every k and ? < 1, we have that \\xfk\\ < K + 1. Because F is an ?-net
for (K + l)BFn, there is 4' G F such that ||x'fc - x'?\\ < e.
O
Fact 8.18
Let (X, || • ||) be a Banach space, and let \\ ¦ \\* be the dual norm of X*. If
|| • ||* is locally uniformly rotund, then || • || is Frechet differentiable.

250 8. Differentiability of Norms
Proof: Let xG5j, and choose / E Sx* such that f(x) — 1. Let fn E Sx*
satisfy lim(/n(#)) = 1. We have
2>||/+/n|r>(/ + /n)(x)-2.
Therefore limB||/n||*2 + 2||/||*2 - ||/ + /„||*2) = 0, and by the LUR
property, lim||/n — /||* = 0. By the Smulian lemma, || • || is Frechet
differentiable.
?
Theorem 8.19 (Kadec; see, e.g., [DGZ3])
Let X be a Banach space. If X* is separable, then X admits an equivalent
Frechet differentiable norm.
Proof: By Proposition 2.9, X is separable. Let {xn} C Sx be dense in Sx,
and let {fn} C Sx* be dense in Sx* • For n E N, put Fn = span{/i,..., /„}.
Define a norm ||| • ||| on X* by
oo oo
Ill/Ill2 = ll/ll2 + E 2_" diSt(/> F"J + E S-"/2^),
where || • || denotes the original dual norm on X*. As in the proof of
Theorem 8.17, we find that ||| • ||| is an equivalent LUR norm on X*. We claim that
I • I is w*-\ower semicontinuous. First, note that Fn + Bx* is ii>*-closed
because Fn is u>*-closed and Bx* is w*-compact by Alaoglu's theorem. Since
{/ E X*; dist(/,Fn) < 1} = Fn + Bx*, the functions dist(/,Fn) are
u;*-lower semicontinuous and so is the supremum of their weighted partial
sums. Consequently, | • || is w*-lower semicontinuous and thus it is the dual
norm to an equivalent norm || • \\l on X by Lemma 8.8. || • ||x is Frechet
differentiable by Fact 8.18.
?
Theorem 8.20 (see, e.g., [DGZ3])
Let X be a Banach space.
(i) If X is separable, then X admits an equivalent norm that is
simultaneously LUR and Gateaux differentiable.
(ii) If X* is separable, then X admits an equivalent norm that is
simultaneously LUR and Frechet differentiable.
Proof: ([JoZ2]) We will prove (ii), the proof of (i) being similar. Let || • ||
be an equivalent LUR norm on the separable space X (Proposition 2.9
and Theorem 8.17). Using the method of proof of Theorem 8.19, we define
equivalent dual norms || • ||* on X* by
^ oo oo
i2 = di/ir J + - (E2_i dist(/> fo2 + E2-i/2 (**-)) •
i=l i=l

8. Differentiability of Norms 251
Then || • ||* are LUR and their predual norms || • ||n converge to || • || uni-
oo
formly on bounded sets. Define a norm || • ||0 on X by ||x||q = J2 ^~n\\x\\n
n = l
for x G X. Then || • ||0 is an equivalent Frechet differentiable norm on X.
The differentiability follows by a standard argument because the derivatives
of norms || • ||n on X are uniformly bounded and || • || is Frechet
differentiable at 0 (Exercise 8.24). We will check that || • ||0 is LUR. To this end, let
lim B||xjb||g + 2||s||§- ||z + ?'n||o) = 0. Then {xk} is bounded and a similar
k —k»
limit relation is true for each norm || • ||n, as shown in Theorem 8.17. Since
the norms || • ||n converge uniformly on bounded sets to || • ||, we get by the
standard limit interchanging rule that lim B||?||2+2||?jb||2 — ||ar+^fc||2) = 0.
k —+ oo
Because || • || is LUR, we get \\x — Xk\\ —> 0.
D
The method of combining norms with rotundity and smoothness
properties is called the Asplund averaging (see, e.g., [DGZ3]).
If X is a Banach space with a separable dual, then the set of all equivalent
Frechet differentiable norms on X is residual in the set of all equivalent
norms in the topology of uniform convergence on the unit ball ([DGZ3]).
However, it is not Gs there if the space is infinite-dimensional ([BGK]).
Theorem 8.21 (Asplund [Asp], Lindenstrauss [Lin2])
Let X be a Banach space such that X* is separable. If f is a continuous
convex function on X, then the set of all points of Frechet differentiability
of f is a dense Gs set in X.
In the proof, we will use the following fact.
Lemma 8.22
Let f be a convex function on an open convex set U in a Banach space X.
If f is bounded above, then X is locally Lipschitz on U. In particular, every
continuous convex function defined on an open convex set C in a Banach
space X is locally Lipschitz on C.
PROOF: We will first show that if / is bounded above on B(xo,r), then
it is bounded on B(xo,r). Assume without loss of generality that xq — 0
and that / is bounded above by K on the closed unit ball Bx- By the
convexity of /, we have that /@) = f(\{x + (—x))) < \f{x) + \f(~~x) f°r
every xeBx. Thus f(x) > 2/@) - f(-x) > 2/@) - K.
The lemma now follows from the following claim.
Claim
If a convex function f is bounded by 1 on Bx, then f is Lipschitz with the
constant 5 on \Bx-
Indeed, assume that x,y G \Bx and f(y) — f{x) > 5||y— x\\. Consider
y — x
the point z = y+ -zr. rr- Clearly, z ? Bx - On the other hand, since the
2\\y-x\\

252 8. Differentiability of Norms
points ?, y, z lie on a line in this order, by the convexity of / we have
/(^)-/(y)>/(y)-/(»). *
—n n ~ —n n— > °*
If-2/11 \\y -x\\
Also, \\z-y\\ - \. Therefore f(z) > f(y)+\ > -1+f = |, a contradiction.
?
Proof of Theorem 8.21: (Preiss-Zajicek) Let A be the set of all points
in X where / is not Frechet differentiate. We will show that A is of first
category in X. Consider the epigraph {(x} y) G X 0 R; y > f(x)} of /.
Using the separation theorem in X 0 R (note that its dual is isomorphic
to X* 0 R), for every x G X we find a functional px G X* such that
f(x+h) — f(x) > px(h) for every h G X. Since / is not Frechet differentiable
at points of A, for every x E A we find mx G N such that
f(x + h)-f(x)-pF{h) ^ 1
hmsup TTTj ^— > .
h-+o \\h\\ rnx
For m G N, put Am = {x G A] mx = m). Given m G N, consider the cover
of X* by all open balls in X* of radius ^^- Since X* is separable, by the
Lindelof property, let {B™}k be a countable subfamily of these balls that
covers X*. For k G N, define j4m>jb = {z G Am; p* G Bjf1}.
We have A — (J j4mjjfc. Hence it is enough to show that Amj. is nowhere
dense for each m,k. Fix m and ?, then choose any x G Am^ and a
neighborhood U oi x. We will show that there is a point y G U that has a
neighborhood V such that V C\ Am)k = 0.
By Lemma 8.22, assume that U is of the form U — 5j(x, r), the open
r-ball centered at #, where r is chosen so that / is Lipschitz with constant
K > 1/ra on Bx(Xi r). Since x G ^4m> there is h G X, ||ft|| < r such that
f(x + h)-f(x)>WL + if(h). A)
We will show that B%(x + ft, ||/i||/12ifra) H Am)jfc = 0. Assume that there
is z G Bx(x + h, ||/i||/12ifm)nAmfjb. Since z G Am^ and x G ^m,fc, by the
definition of Am^ we obtain \\px — pz\\ < j^. By the choice of pzy
f{x)-f{z)>tf(x-z). B)
Adding A) and B), we have
f(x + h)-f(z) > f(x-z)+^+px(h)
= px(x + h-z) + (p*-px){x-z) + ^.
C)
INI
2m'
Since \\x+h- z\\ < J^L and ||p*|| < R, we have \px (x -\- h — z)\ < |
Furthermore, \\z - *|| < \\z - (x + h) + A|| < j^L + ||/i|| < 2||/i|| and
\\PX~P2\\ < Tib Therefore \{f - px)(x - y)\ < ^ • 2||A|| = M. Hence,

8. Differentiability of Norms 253
from C) we obtain
m 6m 3m 2m
This contradicts the fact that |||ar + h\\ - \\z\\\ < \\x + h - z\\ < J^L.
Therefore, / is Frechet differentiate on a residual set in X. The fact that
it is actually Frechet differentiable on a dense Gs set in X follows from the
following general lemma.
Lemma 8.23
Let f be a continuous convex function on a Banach space X. Then the set
G of all points in X where f is Frechet differentiable (possibly empty) is a
Gs set in X.
Proof: For n G N, define
Gn = h e X; inf sup /(« + '») + /(«-fr)-2/(*) < I).
Since / is convex, Ar+ y)+f\*- y)~ f\x) js decreasing as 6 \ 0+, and we
get G = f]Gn. Hence it suffices to show that each Gn is an open subset of
X. To this end, let x G Gn and / be X-Lipschitz on B°(x, a). There exists
C < - and 6 < % such that
n 2,
mn f(x + 6y)+f(x-Sy)-2f(x)
sup < G.
y?Sx °
Choose 0 <e < min{|, jl(^-C)}. We claim that B°{x,e) C Gn. Indeed,
given z G B(x,e), we have for all y G Sx'-
f(z + 6y) + f(z-6y)-2f(x)
8
< f(x + by) + L\\z - x\\ + f(x - 6y) + L\\z - x\\ - 2f(x) + 2L\\z - x\\
<C+4L?<c+(i_c) = i
?
We can summarize some of the results in this section.
Theorem 8.24
Let X be a separable Banach space. The following are equivalent:
(i) X* is separable.
(ii) X admits an equivalent Frechet differentiable norm.
(Hi) X admits an equivalent norm that is simultaneously LUR and Frechet
differentiable.
(iv) For every continuous convex function f on X, the set of all points of
Frechet differentiability of f is a dense G$ set in X.

254 8. Differentiability of Norms
The situation for non-complete normed spaces is different. In fact, every
normed space with a countable Hamel basis admits an equivalent Frechet
differentiable norm ([Vanl], [Hajl]). It is not known whether X* is
separable whenever a separable Banach space X admits an equivalent norm whose
restriction to every closed subspace has a point of Frechet differentiability.
Definition 8.25
A Banach space X is called an Asplund space if for every convex continuous
function f on X, the set of all points of Frechet differentiability of f is a
dense G$ set in X.
Thus by Theorem 8.21, every Banach space whose dual is separable is
an Asplund space.
Preiss proved that if X is an Asplund space, then every Lipschitz function
on X is Frechet differentiable on a dense set in X ([Pre]). However, it is not
known whether a common point of Frechet differentiability can be found
for two Lipschitz functions on an Asplund space. For more information in
this direction, we refer to [BeLi].
Concerning the Gs property in Lemma 8.23, we note that there exists
a Lipschitz function / on R such that the set of points where / is not
differentiable is a residual set in R ([Zah]).
Theorem 8.26 (see, e.g., [DGZ3])
Let X be a Banach space. The following are equivalent:
(i) X is an Asplund space.
(n) For every convex continuous function defined on a convex open subset
C of X7 the set of all points of Frechet differentiability of f is a dense Gs
set in C.
(iii) Every equivalent norm on X is Frechet differentiable at some point of
X.
(iv) Every separable closed subspace of X has a separable dual
We only prove (iv) => (i) and refer to [DGZ3] for the rest.
Proof: Assume that / is a continuous convex function on X such that the
set D of Frechet differentiability points of / is not dense in X. Thus there
is an open ball B in which there is no point of Frechet differentiability of
/. For n G N, let
Gn = L e B; inf sup K* + W + H*'W ~W*) < I}.
v. 6>0y€Sx * n'
As in the proof of Lemma 8.23, Gn are open for all n, and from the convexity
of / we have that D f) B = f] Gn. Since B is a Baire space, we have that
Gn is not dense in B for some n. Thus there is n and an open ball U in
B such that Gn C\ U = 0. We will find a separable closed subspace Y of X
such that no point of Frechet differentiability of the restriction of / to Y
is in U. This will contradict Theorem 8.21.

8. Differentiability of Norms 255
The desired subspace Y is constructed as follows. Choose x\ E U.
Because x\ (? Gn, from the monotonicity of the quotients we have that there
is a sequence {yj}j in Sx such that, for all 6 > 0,
f(x1+6y}) + f(x1-6y})-2f(x1) 1
sup J- l > —.
j o In
Let Y\ be the closed linear span of x\ and {y]}j. If an increasing sequence of
separable closed subspaces Yi, Y2, • • •, Yk has been constructed, let {xkyi}i
be a countable dense subset of Yk H U and for each i choose a sequence
{y)}j in Sx such that, for all 6 > 0,
f(xk}i + 6$) + f(xkii - 6y)) - 2/(sM) _1_
j 5 ~ In
Let y^+i be the closed linear span of Yk and {xk,i}iV{yj}i,j> Let Y = |J^-
Then {xkti}k,i is dense in U fl Y. Moreover, each x^; is not in (?2n5 where
the latter set is the (?2n defined for the restriction of / to Y. Because
Gin are open in Y, we have that U O Y is disjoint from C?2n and thus the
restriction of / to Y is Frechet differentiable at no point of U Pi Y (see
Exercise 8.16).
?
Let X be a Banach space such that X* is separable, and let Y be a
closed subspace of X. Given an equivalent Frechet differentiable norm on
Y, it is not known whether it can be extended to an equivalent Frechet
differentiable norm on X. For Gateaux differentiability, the answer is known
to be negative; see, e.g., [DGZ3].
Extremal Structure
Definition 8.27
Let C be a convex set in a Banach space X. A point x E C is called an
exposed point of C if there is f E X* such that f(x) — supc(/) and for
every x1 E C with f(x') — supc(/) we have x — x'.
A point x E C is called a strongly exposed point of C if there exists f E X*
such that f(x) — supc(/) and xn —» x for all sequences {xn} C C such
that lim(/(a?n)) = supc(/).
Theorem 8.28 (Lindenstrauss [Lin2], Troyanski [Trol])
Let C be a weakly compact convex set in a Banach space X. Then C is the
closed convex hull of its strongly exposed points.
We will give a proof for X separable; for the general case, see [Lin2] and
[Trol].

256 8. Differentiability of Norms
Proof: Assume without loss of generality that 0 G C C Bx- Define a
function F on X* by F(f) = sup{/(#); x G C}. Then F is a
continuous convex function on X* (the continuity follows from the boundedness
of C). Given / G X*, by the weak compactness of C, there is z G C
such that f(z) = sup {/(#); ? G C}. Then for every # G X* we have
F(<?) = supc(<?) > g(z\ and hence F(g)-F(f) = F(g)-f(z) > (g-f)(z).
Therefore, z can play the role of pf in the Preiss-Zajicek proof of
Theorem 8.21. Since these z lie in a separable space X, following that proof we
obtain that F is Frechet differentiate on a dense G§ set D C X*. Also, if
F is differentiate at / and f(x) — supc(/), then Ff(f)(g) = g{x) for every
g G X*, since both the derivative and x give a supporting hyperplane to the
epigraph at the point (/, F(f)), which is unique due to the differentiability
oiF.
We claim that if F is Frechet differentiate at / and f(x) = supc(/),
then x is a strongly exposed point of C. Let xn,yn G C be such that
f(xn) —> suPc(/) an^ f(Un) —> suPc(/)- We must show that xn —* x. It
suffices to show that \\xn — yn\\ —> 0 (we then apply this to yn = x). From
the Frechet differentiability of F at /, given e > 0 there is 8 > 0 such that
F(f+h) + F(f-h)-2F(f) < e\\h\\ for every h G X* with ||/t|| < 6. Choose
no such that for n > no we have max{|/(^n) — F(f)\, \f(yn) — F(f)\} < ?<$•
Then, whenever n > no and \\h\\ < <5, we have
h(xn - yn) - (/ + n)(zn) + (/ - fc)(t/n) ~ f(x*) ~ f(Vn)
< F(f + h) + F{f-h)-f(xn)-f(yn)
< 2F(f)-f(xn)-f(th) + e\\h\\
= (**(/) " /(*n)) + (**(/) - /fan)) + e||/l|| < 3rf,
so we have \\xn — yn \\ < 3e for every n > no. This shows that # is a strongly
exposed point of C, exposed by /.
Let S be the set of all strongly exposed points of C. We now show that
conv(S) = C. Assume the contrary. By the separation theorem, there is
/ G X* and some 8 > 0 such that sup5(/) + 8 < supc(/). Since the points
of the Frechet differentiability of F form a dense set in X*, we can assume
without loss of generality that / is a point of Frechet differentiability of
F. Let f(x) = supc(/). We already observed that ? is a strongly exposed
point and thus sup5(/) > supc(/), a contradiction.
?
A modification of this proof gives the following result.
Theorem 8.29 (Lindenstrauss [Lin2], Namioka, Phelps [NaPh])
Let X be a Banach space such that X* is separable. If C is a bounded closed
convex subset of X*, then C is the closed convex hull of its strongly exposed
points.
In the proof, we will use some ideas from [CoEd].

8. Differentiability of Norms 257
Lemma 8.30
Let X be a Banach space. For every x ? X \ {0} and a G X** \ {0}; there
is an isomorphism T of X* onto X* such that T*(x) — a.
PROOF: Consider the subspace Z — span{#,a} of X**. Working in the
u>*-topology, we get that Z is complemented in X** by a w*-continuous
projection P. Let A be a one-to-one linear operator on Z that transfers x
to a. Let G be an operator on X** denned by G(x**) = (Ix** — P)(x**) -f
A(P(x**)). Then G is an isomorphism of X** onto X** that is w*-w*-
continuous, so G — T* and T is the desired isomorphism.
?
Lemma 8.31
Let X be a Banach space such that X* is separable. Let C be a bounded
closed convex subset X* such that 0 G C. Let T be an isomorphism of
X* onto X*. Define a function F on X** by F(x**) = supc(x**) and a
function H on X by H(x) = sup (x) (x is considered a function on X*). If
H is Frechet differentiable at xo, then F is Frechet differentiate at T*(xq).
Proof: First, note that F and H are convex functions that are bounded
on bounded sets and thus they are continuous (see, e.g., Lemma 8.22). Let
/o G T(C) be such that fo(xo) = sup (xq). By Smulian's lemma, it
follows that /o is the Frechet derivative of H at xo and /o is actually in T(C)
(if fn G T(C) are such that fn(xo) —»• fo(xo), then {/n} is norm-convergent
to /o by Smulian's lemma (Chapter VIII of [DGZ3]), and necessarily goes
to /o). This means that /o(#o) = sup (xq) = sup(T*(xo)).
T(C) C
Let /o = T(go) for some go G C. Whenever gn G C are such that
^*(*o)(<7n) -+T*(z0)(<7o) =sup(T*(z0)) = sup(^o) = /0(a?o),
C T(C)
then
T(gn)(x0) = T*(x0)(gn) -* /o(*o) - T@O)(*o).
Since fo is a point of T(C) strongly exposed by #o, we have that T(gn) —>
T(go) = /o in norm. T is an isomorphism, so we have that #n —> #0 in
norm. This means that go is strongly exposed in C by T*(xo) and thus F
is Frechet differentiable at T*(xq) with the derivative go.
D
Proof of Theorem 8.29: Consider F on X** as in Lemma 8.31. We
will show that F is Frechet differentiable on a dense set in X**. Given
a G X** \ {0}, fix some x0 G X \ {0} and find an isomorphism T of X*
such that T*(xo) = a by Lemma 8.30. For this, T define the function H

258 8. Differentiability of Norms
as in Lemma 8.31. Since X* is separable, arbitrarily close to xo is a point
x\ of Frechet differentiability of the function H on X. Then F is Frechet
differentiate at T*(x\), which is arbitrarily close to T*(xq) = a. The rest
of the proof is similar to that of Theorem 8.28.
?
Theorem 8.32 ([CoLi], [GoKa])
Let K be a weakly compact set in a Banach space X.
intersection of a family of finite unions of balls.
Then K is the
Proof: We will present the Corson-Lindenstrauss proof of their version of
this theorem, when X is separable and reflexive. Let r be the topology on
Bx in which the closed sets are exactly the intersections of finite unions of
balls in X. Note that r is weaker than the weak topology on Bx- Because
Bx is weakly compact, to prove Theorem 8.32, it is enough to show that
r is Hausdorff since r and the weak topology then must coincide on Bx •
Let yi,y2 G Bx, Vi / 2/2- It is enough to find balls B(xur1), B(x2,r2)
in X such that yi (fc B(x(,ri), i — 1,2 and B(xi,ri) U B{x2)r2) D Bx-
Indeed, then Bx \ 5(x2-,rz), i — 1,2, are two disjoint open sets in r and
Vi G Bx \ B(xi,ri) for 2 = 1,2, showing that r is a Hausdorff topology.
Put z — y\ — y2 and find a point u G Sx of the Frechet differentiability
of the norm of X such that \\u •
M1
< |. Indeed, X* is separable, so we
can use Theorem 8.21. For n G N, we define x™ — y2 — (n — ^\\z\\)u and
Vi + (n-
Pi -2/1II =
>
|z||)u. We then have
'+ (n-
z u
nu + -^-u — \\z\\u-\- z
(„+ji|ii)„|-ll,
z-f nu — 7j\\z\\u
nu + ^u - (\\z\\u - z)
>n+M_M>n.
M
Similarly, we show that \\x2 — y2\\ > n for all n G N, so yi ? B(xfin) for
i= 1,2.
The proof will be complete when we show that B (x™, n) \JB{x^ , n) D Bx
for some n G N.
Suppose this is not the case. Then, for every n G N, there is zn G Bx
such that \\zn - y2 + (n - f||z||)u|| > n and ||zn — t/i — (n — | ||z||)ii|| > n.
Thus, for all n G N,
^+~(zn-y2.--||z||^)
77- O
>l,
(l)
l
(m
\u)
> l.
B)
n x" 3
Let u* be the Frechet derivative of the norm of X at u. Then, from the
Frechet differentiability, we have \\u + y\\ = 1-\-u*(y) + o(\\y\\) and thus by

8. Differentiability of Norms 259
A) and B), and using the fact that {zn} C Bx for n E N, we have
IuV-W-§||*||«) = |u+I(*»-jfc-§||z||u)|-l-0(I)
> MS)
and similarly ±u* (yx - zn - §||z||u) > -o(?).
By adding the latter two inequalities, we obtain
Lu*(yi-y2-l\\z\\u)>-o(±).
Given e > 0, there is no such that —o(~) > — ^ for n > no- Thus for n > no
we have ^u* (yi - y2 - %\\z\\u) > -| and hence tz*(?/i - ?/2 - f||z||u) > -?.
Since this holds for all ? > 0, we have u* (yi — ?/2 — |lkllw) > 0- Hence
u"(z) - ±\\z\\ = u*(z) - §||z||u» = «* B/! - J/2 - !||*||«) > 0.
Thus u*(z) > 11 \z\ |, which contradicts the fact that ||u*|| = 1.
D
Theorem 8.33 (Mazur, Phelps; see, e.g., [DGZ3])
Let X be a Banach space whose norm is Frechet differentiable. If C is a
bounded closed convex subset of X, then C is an intersection of balls in X.
Proof: Assume that 0 ? C. We will find a ball B(x,p) such that C C
B(x,p) and 0 g B(x,p).
Using the separation theorem, we find /o E Sx* such that inf(/o) > 0.
By the Bishop-Phelps theorem, we may assume that /o — ||?o||' for some
xo e SX- Put e - |inf(/) and Bn = B(nex0,(n- l)e).
Note that 0 ^ Bn for every n > 2. Hence it suffices to show that C C Bn
for some n > 2.
Assume that for every n > 2 there is xn E C\Bn. Then \\xn — nexo\\ >
(n — \)s and thus \\xq — ^xn\\ > 1— ^ for every n > 2. The norm is Frechet
differentiable and /o = H^oll', so for h E X we have ||xo+n|| — ||a?o|| —/oCO =
r(h), where lim tiW- = 0. Using this for h — —~xn, we obtain
r(-±xn) =||xo- —Xnll-l + Zof—a?n) > -^ + ^ = 1.
Hence for n > 2 we have
r(~^xn) > e > e
11 l xn\\ - \\xn\\ ~ sup||zn||'
i ne
Because {xn} is bounded, ^jxn —* 0 and we obtained a contradiction with
lim ^ = 0. Therefore, Bn D C for some n.
?
The following theorem shows another application of smoothness.

260 8. Differentiability of Norms
Proposition 8.34
Let X be a Banach space with a Frechet differentiable norm || • ||. Let {e2}
be a Schauder basis of X with associated projections Pn. 7/lim||Pn|| = 1,
then {e{} is shrinking.
Proof: Let x* ? Sx*- Given e > 0, by the Bishop-Phelps theorem find
a support point u* ? Sx* (supported by some u ? Sx) such that \\u* —
#*|| < e. Since || • || is Frechet differentiable, there exists 0 < 6 < e such
that diam(S(M)) < e/2, where S(u,6) = {y* E Bx*\ u(y*) > 1-6}.
Observe that \\P?\\ —> 1 and that jF?(t/*) ^ u*. Choose n0 ? N such
that for every n > n0, \P*{u*)(u) - 1| < f and ||P*(u*)|| < 1 + f • Let
6* G#x*, ||fr*-^(«*)||< f. Then \b*(u) - 1| < 5, so ||6*-ti*|| < | and
||P*(ii*) — w*|| < e for every n > no. This shows that {e2} is shrinking.
?
We now discuss norm-attaining operators. We introduce the following
notation. Let X, Y be Banach spaces and T ? B(X, Y). For a bounded set
C in X, we put ||T||C = sup{||T(ar)||; x ? C}. If there is c ? C such that
||T(c)|| = II^Hc w^ say that T attains its supremum over C.
Theorem 8.35 (Lindenstrauss [Lin2])
Let X,Y be Banach spaces. If C is a w*-compact subset of X*, then the
set of all dual bounded linear operators from X* to Y* that attain their
suprema over C is dense in the Banach space of all dual bounded linear
operators from B(X*, Y*).
In the proof, we will use the following lemma.
Lemma 8.36
Let T*: X* —+ Y* be a dual bounded linear operator and C be a w*-compact
subset of X*. The following are equivalent:
(i) T* attains its supremum over C.
(ii) There exist {fk} C C and {%•} C SY such that \T*(fk)(yj)\ > \\T*\\C-
j for allj,keN, k >j.
Proof: Assume that (ii) holds and / is a w*-limit point of {fk}- Since
T* is w*-w*-continuous, we get that \T*(f)(yj)\ > \\T*\\C - A for every j.
Therefore ||T*(/)|| = ||T*||C.
If ||T*(/)|| = \\T*\\C for some / ? C, put fk = / for k ? N and pick
K e SY, j ? N such that |T*(/)(W)| > ||T*(/)|| - i.
D
Proof of Theorem 8.35: Let T*:X* -> Y* be a norm-one dual
bounded linear operator, and let C C Bx* be a uAcompact set. We put
c = ||T*||c < 1 and assume that c> 0.

8. Differentiability of Norms 261
Take any e G @,min{^, |}) and choose a decreasing sequence {sk^-i
oo oo
of positive numbers such that 2 ]P a < e, 2 Y2 ei < el an(^ 2?|c-fl4?jb <
»=i «=fe+i
^- for every k. We set T-j" = T*. By induction, having constructed a dual
operator T* G B(X\Y% find /* G C such that ||7?(/jfc)|| > ||T*||C - e\c
and yk G 5y so that T?(fk)(yk) > H^jfe (/jb)|| -e|c, and finally define a dual
operator T?+1 by
2*+i(/) = W) +ekTk{f){yk)T*k{fk).
We claim that {T? } satisfy the following properties:
A) | < \\T*k\\ < I and §c < II^Hc < |cfor all *, ||^-^||c < 2 ? ?,-c,
»=i
and ||T/ - IJII < 2 fft for j < i;
B) \\n+1\\c > WWc + ek\\Tn\2c ~ 4^c for all k;
C) ||3THc> ||2;||c >c for j<t;
D) |77(A)(yi)| > 117711c " 14?,- for j < *.
Indeed, A) follows by induction. To see B), write
\\n+i\\c > \\n+i(fk)\\ = \\(i+skn(fk)(yk))n(fk)\\
> \\n(fk)\\(l + ek(\\n(fk)\\-elc))
> {\\Tk\\c-elc){\ + ek{\\T*k\\c-2elc))
= m\\c - ele + ek\\Tt\\2c ~ Hc\\n\\c + 24c2
> \\n\\c + ekm\\2c-4elc.
C) follows from A) and B). To see D), write
ii27+i(/*)ii > m(fr)\\-m-T*+1\\c
fc-1
> ||I* He - e\c - 2 J2 ^c > WTUi\\o " 3e?c
Using this, B), and the definition of X?+1, we have for j < k:
e,-|37(A)(%-)l 11271k + 112711c > l|77+1(/OII>H27+illc-3e2c
> \\T*\\c + ej\\T*fc-4e*c-Ze2jc
= ||^||c + ?j||^|lc-7e?c.
Therefore,
|27(/*)(y>)| > ||27||c - 7ejc/\\T;\\c > ||27||c - 14?>.
It follows from A) that {T;*} converges in norm to some operator T G
B(X*,Y*) satisfying ||T*-f || < e and ||f-T;||c < e^c for every j. Since
Tj* are w* —w* continuous, we obtain that T is t^-t/Acontinuous on Bx* •

262 8. Differentiability of Norms
By the Banach-Dieudonne theorem, this means that T is u>*-u>*-continuous
on X*, so T is a dual operator.
If 1 < j < &, then, according to D),
\f(h)(yj)\ > \T*{h){yj)\-\\T*-f\\c ^
> ||i;q|c-i4?j.-?|_lC>||f||c-i.
By Lemma 8.36, this means that ||T||c is attained at some /EC.
?
Let T be a bounded linear operator from a Banach space X into a Banach
space Y. We say that T attains its norm if there is xq ? Bx such that
||T|| — ||T(?o)||- Then we have the following consequence of Theorem 8.35.
Corollary 8.37
Let X, Y be Banach spaces. If X is reflexive, then the set of all operators
in B(X,Y) that attain their norm is dense in B(X,Y).
The proof of the preceding theorem together with Kadec's renorming
theorem for separable Banach spaces by locally uniformly rotund norms
gives an alternative proof of a part of Theorem 8.28.
Corollary 8.38 (Lindenstrauss)
Let X be a separable Banach space. If C is a convex weakly compact set in
X, then C has a strongly exposed point.
Proof: By Theorem 8.17, we may assume that the norm of X is locally
uniformly rotund. Let / be the identity operator on X. Arbitrarily close to
i" is an operator T on X (that is necessarily an isomorphism, Lemma 7.15)
that attains its supremum over C at some xo G C. Let / G Sx* be such
that f(T(xQ)) = ||T(z0)||. We claim that T*(f) is strongly exposing xo in
C. Indeed, if x € C, then ||T(x)|| < ||T(x0)|| and thus f(T(x)) < \\T(x)\\ <
\\T(x0)\\ = /(T(*0)). Therefore alsoT*(/)(x0) = sup(T*(/)(x)).Ifxn € C
xec
have the property that T*(f)(xn) -> T*(/)(x0), then
2||T(x0)|| > \\T(xn + xo)\\>f(T(xn + x0))=T*(f)(xn + x0)
-v 2r(/)(x0) = 2/(r(xo))=2||T(x0)||.
By the local uniform rotundity, we have ||T(xn) — T(a?o)|| —^ 0, and since
T is an isomorphism, we have \\xn — xq\\ —>¦ 0.
D

8. Differentiability of Norms 263
Exercises
8.1 Let / be a convex function on a Banach space X. Show that if / is
bounded above on some ball, then / is continuous on X.
Hint: Assume / < K on xo -f 8Bx • For h E \Bx, we have f(h) < \f(x$ +
2h) + !/(-a?o) < f + |/(-«o) = i- Let a: E X. For y E 5 = x + \BX, we
get /(y) < |/B(i/ - x)) + |/Bar) < L\f{2x). Thus / is bounded above
on B\ hence by Lemma 8.22, / is continuous at x.
8.2 Let / be a finite convex function on a finite-dimensional Banach space
X. Show that / is continuous on X.
Hint: / is bounded above on the symmetric convex hull of the basis vectors,
which contains the origin as an interior point. Then use Exercise 8.1.
8.3 Show that a finite lower semicontinuous convex function / that is
defined on a whole Banach space must be continuous.
Hint: Use the Baire category theorem.
8.4 Let U be a convex subset of a vector space X) /:[/—* R. Show that /
is convex if and only if the function t \—> -^ y)-J\x) js increasing in t for
all x E U, y E V and all t for which x -\-ty E U.
Hint: Assume / is convex; take x E U, y E V, and 0 < t < s such that
x + (s + i)ye U. Then (x + ty) = Xx + A - X)(x + (t + s)y) for A = ^.
Applying convexity, we obtain /fr+WW < /(*+(«+ff)-/(*).
Given x,y ? U and A E @,1), set z = y — x\ then A# + A — A)y =
x + A — A)z and y — x +\z.
8.5 Prove that every convex function / defined on an open interval I C R
is differentiable at all but (at most) countably many points of /.
Hint: Observe that d+/(x)(l) = lim /(*+*)-/(*), the derivative of / at
x from the right, is a nondecreasing function of x. Prove then that, at
any point where / fails to be differentiate, the monotone function x —»
d+f(x)(l) has a jump. Because there are not more than a countable number
of jumps, the conclusion follows.
8.6 Let /, g be convex functions on a Banach space X such that / > g on X
and f(xo) = g(xo). Assume that / is Frechet (resp. Gateaux) differentiable
at #o- Show that g is Frechet (resp. Gateaux) differentiable at x$.
Hint: Use Lemma 8.3.
8.7 Let /, g be convex continuous functions on a Banach space X and
assume that / is not Frechet differentiable at x E X. Show that / + g is
not Frechet differentiable at x.
Hint: By the assumption, f{x + h) + f(x — h) — 2f(x) > eh.

264 8. Differentiability of Norms
8.8 Show that the norm || • || of a finite-dimensional Banach space is Frechet
differentiable at x if it is Gateaux differentiate at x.
Hint: Smulian's lemma or a direct computation.
8.9 Let a function / be Lipschitz on Rn and Gateaux differentiable at x.
Show that / is Frechet differentiable at x. Is this also true for Lipschitz
maps from Rn to a Banach space Y?
Hint: Proceed by contradiction, using the compactness of the unit ball and
the Lipschitz property of /. The second part: Yes.
8.10 Show that f(x) = ?2sin(l/z) is a Lipschitz function on R which is
differentiable everywhere, yet its derivative is not a continuous function.
Thus, Corollary 8.5 does not work for Lipschitz functions.
8.11 Is it true that a Lipschitz function on R2 is Gateaux differentiable at
#o G R2 if it is differentiable at xq in all directions?
Hint: No. The directional derivative need not be linear; check
f< ^ / Tpfe for (s.j,) #@,0),
f(x,y) - < V* +y
1 0 for (*,!,) = @,0).
8.12 Let Xi be vectors in a Banach space X such that span{x2} — X.
Assume that / is a continuous convex function on X such that, at all
points of X, all directional derivatives in the directions of {xi} exist. Is /
Gateaux differentiable on XI
Hint: Yes, use the uniqueness of the supporting functional, which is
sufficient in the directions of {xi}.
8.13 Let / be a Lipschitz function on a Banach space X and xq G X.
Assume that / is differentiable at xo in a dense set of directions. Show that
/ is differentiable at xo in all directions.
Hint: Show that the differentiation quotients are Cauchy in each direction.
Proceed by contradiction, using the Lipschitz property.
8.14 Let B\ and B2 be the unit balls of two equivalent norms on a reflexive
Banach space and assume that B2 is the ball of a Gateaux differentiable
norm. Show that then B\ + B2 is the ball of a Gateaux differentiable norm.
This fact is behind most first order differentiable renormings.
Hint: B — B\ + B2 is closed because Bi are ^-compact. Let x G B and / G
X* have the property that f(x) = max{f(y)] y G B}. Then x — xi + X2,
where xi G Bi. Note that max (/) = f{x\) -f /(#2) = max(/) + max(/),
so f(x{) — max{/(y); y G Bi}. Since ?2 is the unit ball of a smooth norm,
such / is unique.

8. Differentiability of Norms 265
8.15 Let /, g be convex functions on a Banach space X. The inf convolution
of / and g is defined by (/ o g){x) = inf{/(y) + g(x - y)\ ye X} (it
is defined so that its epigraph is the algebraic sum of the epigraphs of
functions involved).
Assume that X is a reflexive Banach space and f(x) = ||#||i, g{x) = \\x\\2
for some equivalent norms || • ||x and || • ||2 on X such that || • 1^ is Frechet
smooth. Show that then fog is a Frechet-smooth convex function on X.
Hint: By the reflexivity of X, given xo ? X, choose yo ? X such that
fog(xo) = f(yo) + g(x0 -j/0). Then for h ? X:
f o #(zo + h) + / o flf(a?0 - h)-2fo g(x0)
< f(yo) + tf(zo + h-y0) + f(y0)
+g(x0 -h-y0)- 2(/(y0) + #(*o - 2/o))
= 9(xq -yo + h) + g(xQ -y0-h)- 2g(x0 - y0).
From the differentiability of g at (xo — yo)> the differentiability of fog at
#o follows.
8.16 Let / be a continuous convex function defined on an open convex
subset C of a Banach space X. Show that, for every xq ? C, there is a
continuous convex function / defined on X and such that / = / on some
neighborhood of Xq.
Hint: Define / by +oo outside C and consider its inf convolution with the
function (j)n(x) = n\\x\\. Since / is locally Lipschitz, / = f o<j>n close to xo
for n large enough.
8.17 Let X be an infinite-dimensional separable Banach space. Show that
there is a continuous convex function / on X such that / is unbounded on
Bx.
Hint: (Vanderwerff) Let rn be even continuous convex functions such that
rn are nondecreasing on [0,oo), rn = 0 on [0,1/2] and rn(l) = n. Let
fn ? Sx* be such that fn —> 0 (Josefson-Nissenzweig). Consider /(z) =
J2Tn{fn(x))' Note that the sum is locally finite.
8.18 Let X be a Banach space with a Gateaux differentiable norm. Let
/ ? Sx* not attain its norm. Define <p on X by <p(x) = \\x\\2 — f(x) and
M = {x ? X; cp(x) < 0}. Show that M is a closed bounded convex set,
<p is a differentiable convex function such that ^(#) = 0 on the boundary
dM = {x; (p(x) = 0}, but ip1 / 0 on Int(M).
Thus, Rolle's theorem is not true in infinite-dimensional spaces.
Hint: Assume that <pf(xo) = 0 for some a^o ? Int(M). Then (p attains its
minimum at #o by the convexity of (p. Thus xq ^ 0 and 2||xo||||a;o||/ = /¦
Hence / is a multiple of the derivative of the norm, and as such it attains
its norm, a contradiction.

266 8. Differentiability of Norms
8.19 Let X be a Banach space and let / be a continuous convex function on
X* that is tu*-lower semicontinuous. Show that if/ is Frechet differentiable
at x* EX*, then /'(^)Gl
Hint: The derivative, as a uniform limit of quotients in Bx*, is also u>*-lower
semicontinuous. Then use its linearity to see that f'(x*) is a functional that
is w*-continuous on Bx* and apply Theorem 4.44.
8.20 (Godefroy) Let X, Y be Banach spaces such that there is an isometry
T of X* onto Y*. Show that if the dual norm of X* is Frechet differentiate
on a dense set in X*, then T is the dual operator to some isometry S of Y
onto X.
In short, if the dual norm of X* is Frechet differentiable on a dense set
in X*, then X is an isometrically unique predual of X*.
Hint: Let D be the set of \\x*\\' for points x* G Sx* of Frechet
differentiability of the dual norm of X*. Note that D C Sx (Exercise 8.19). We claim
that D = conv™ (D) = Bx** • This follows from the separation theorem
and from the fact that for every x* G Sx* we have sup{/(x*); / G D} — 1
(we use here that the set of the points of Frechet differentiability of the
dual norm is dense in X*). From this and the fact that D C Sx, it follows
that conv(?>) = Bx- Therefore span(D) = X. Since isometry preserves the
points of the Frechet differentiability, we have that an analogous reasoning
can be applied to Y*. Let y G Y be a Frechet derivative at some point
y* G Sy*- Then the functional T(y) is the Frechet derivative of the dual
norm of X* at the point T~l(y*) and as such it is from X. Therefore, T*
maps y into X and T is w*-u>*-continuous. Then (T*|y)* = T.
8.21 Let {rn} be a sequence of all rational numbers in @,1). Define a
oo
function / on @,1) by f(x) = J2 2~n\x — rn\. Show that / is a con-
n = l
vex continuous function on @,1) that is differentiable exactly at irrational
points of @,1).
Hint: Use Lemma 8.3 to see that / is not differentiable at rational points.
For irrational points, the fraction can be made arbitrarily small, first
working with the tail and then with the remaining terms.
8.22 Define a map <j> from [0,1] into Li[0,1] by <j>{t) — X[o,t]- Show that <f>
is Lipschitz and nowhere differentiable.
Hint: Differential quotients are not Cauchy.
8.23 Let A be a subset of R with X(A) — 0 (Lebesgue measure). Follow
the hint to show that there is a Lipschitz function / on R not differentiable
at points of A ([Zah]).
Since there is a residual set of measure 0 in R, this shows that the set of
differentiability points of a Lipschitz function need not be residual.

8. Differentiability of Norms 267
Hint: ([BeLi]) Take open sets Gx D G2 DD ... D A such that A(Gn) < 2~n
and A ((a, 6) C\ Gn+i) < ^p for every component (a, 6) of Gn. Put /n(a?) =
A((-oo, x) fl Gn) and / = ?(-l)n+1/- Then / is Lipschitz.
If x G f]Gn, then / is not differentiable at x. Indeed, let (c*n,/?n) be
components of Gn containing x. Fix n, define 7j = ^ _g , and observe
that {yj} is decreasing, 71 = ... = jn — 1, and 7n+i < |. Thus if n is
even, then /%nn)I^n°fn) = 1 - 1 + ... - 1 + 7n+i - 7n+2 + ... < §, while if
Tl is Odd, !{P?j}"n) = 1 - 1 + . . . + 1 - 7n + l + Tn+2 -...>§-
8.24 Let (X, || • ||) be a Banach space. Show that || • || is Frechet
differentiable at 0 and (|| • ||2)'@) = 0.
Let (H, || • ||) be a Hilbert space H. Show that || • || is Frechet
differentiable at every point of H.
Hint: The first part follows by direct calculation.
F(h) — 2(x, h) in the notation of Definition 8.1.
8.25 Show that the supremum norm of cq is Frechet differentiable at x =
(x{) E SCo if and only if ||z|| = max \x{\ is attained at exactly one i.
Hint: If the condition is satisfied, lk+^ll+ll^-^l|-2^1| = 0 for t small enough
and every h G SCo- If this condition fails and, say x\ — x2 — 1, then t\
and e2 are two support functionals in l\ to BCo at x. Note that this means
that the norm of cq is Gateaux differentiable at x if and only if it is Frechet
differentiable at x.
8.26 Show that the canonical norm of l\ is nowhere Frechet differentiable
and is Gateaux differentiable at x — (x{) iff X{ ^ 0 for every i.
If T is uncountable, show that the canonical norm of -^i(r) is not Gateaux
differentiable at any point.
Hint: Let x G Sx- Given e > 0, find i such that \x{\ < e/2 and consider
h = eei. Show that \\x ± h\\ > 1 + e/2 and use Lemma 8.3.
Note that every vector in -^i(r) has a countable support, choose a
standard unit vector outside this support, and use Lemma 8.3.
8.27 Find an example of a Gateaux differentiable norm on a Banach space
that is not Frechet differentiable at some points.
Hint: Any equivalent renorming by a Gateaux differentiable norm of t\
(Theorem 8.13) satisfies the requirement (Theorem 8.26).
8.28 Show that the norm of G[0,1] is nowhere Frechet differentiable.
Show that the norm of G[0,1] is Gateaux differentiable at x G Sc[o,i] if
and only if \x\ attains its maximum at exactly one point of [0,1].
Hint: Note that the distance between two different Dirac measures in
G[0,1]* is two. Given x G 5c[o,i]) choose 2q € [0,1] such that x(to) = 1.

268 8. Differentiability of Norms
Then choose tn ^ to such that x(tn) —» 1. By the Smulian lemma, x is not
a point of Frechet differentiability of the supremum norm on C[0,1].
For the second part, assume that x G Sc[o,i] 1S sucn that x(to) = 1 and
\x(t)\ < 1 for every t ± t0. Put H = {/ G C[0,1]*; ||/|| < 1, f(x) = 1}. If
Hr\Bc[oti]* ^ {?*0}> ^hen this intersection would have at least two extreme
points that would be extreme points of 5c[o,i]*- All the extreme points of
#c[o,i]* are i Dirac measures (Lemma 3.42).
8.29 Let p G (l,oo). Show that the norm of Lp[0,1] is Gateaux
differentiable and calculate its Gateaux derivative.
Hint: By the standard rules (use the monotonicity in the differential
quotient), we get || • \\'x(h) = Wx^'P f ^(t)^-1 sign(x(t))h(t) dt. The
convergence of the integral follows from Holder's inequality.
8.30 Show that the norm of i^ is Frechet differentiable on a dense set in
Hint: Show that the norm is Frechet differentiable at all points x G S^
such that \xi\ — 1 for some i and sup{|?j|; j ^ i} < 1. Such points are
dense in Si^.
8.31 Show that the norm of Loo[0,1] is nowhere Gateaux differentiable.
Thus, Loo [0,1] is not isometric to loo because the norm of ^oo is Frechet
differentiable on a dense set (see the previous exercise). However, these two
spaces are isomorphic (Exercise 6.18).
Hint: Let ||/|| = 1. Assume that there is a sequence of pairwise disjoint
sets In of positive measure such that |/(/) — 1| < ^ for alH G In- Consider
functionals Fn(g) = //(/n)_1 fT f d\i. Let H and G be it>*-cluster points of
{F2n} and {i^n+i}- They give two supporting hyperplanes to the unit ball
at /. As checked on the characteristic function of the union of {hn}, H
and G are distinct.
8.32 Let || • 11^ denote the canonical of l^ and set p(x) = limsup \x{\.
Define |^| = ||#||oo + p(x) for x G ^oo- Show that ||| • ||| is nowhere Gateaux
differentiable.
Hint: It is enough to show that p is nowhere differentiable. If x — (x2) G ^oo
and xnk —> 1 = p{x)) consider the direction h = YK~^)kenk-
8.33 Show that the restriction of the canonical norm of l\ to any two-
dimensional subspace of l\ is not Gateaux differentiable.
Hint: Let a =. (a;) G l\ and b = (&;) G ^i, b ^ 0. Define a function for
A G K by /(A) = ]T \ai -f A6t-1. Show that / is not differentiable for some A.
8.34 Show that l\ has two-dimensional subspaces that are strictly
convex in the canonical norm of t\. This idea of Lindenstrauss ([Lin3]) was
extended by Fonf and Kadec, who found infinite-dimensional u;*-closed
subspaces of l\ with the same property ([FoKa]).

8. Differentiability of Norms 269
Hint: Let {a;} and {&;} be such that j^1} is dense in R. Show that the
function /(A) = ^2\ai + ^i\ is then not affine on any non-degenerate
interval in R.
8.35 Show that cq contains no two-dimensional subspace on which the
standard norm is Gateaux differentiate.
Hint: Then some quotient Q of l\ would have uncountably many extreme
points since the dual to a two-dimensional smooth space is strictly convex.
Every point of the sphere of Q that is identified with the restriction to
the two-dimensional subspace in question extends to an extremal point of
the sphere in l\ by the Krein-Milman theorem, considering the face of all
the extensions. Thus, there are uncountably many such extreme points of
the ball of the standard norm in t\, which is a contradiction because the
extreme points of the ball of t\ are exactly ±e;.
8.36 Let X be a reflexive strictly convex Banach space, and C a closed
convex set in X. Show that there is a unique nearest point to x in C.
Hint: x has a nearest point because X is reflexive (Exercise 3.104). Assume
that x = 0 and dist(z,C) = 1. Let c\ ^ C2 G C satisfy ||ci|| = ||c2|| = 1.
Then |(ci -f C2) G C, yet ||^(ci + C2)|| < 1 by the strict convexity, a
contradiction.
8.37 Let C be a convex closed set in a reflexive Banach space whose norm
is locally uniformly convex. To every x G X assign p(x), the closest point
of C to x. Show that p is continuous.
Hint: Let dist@, C) — 1 and xn —> 0. Let y = p@) and yn — p(xn). Then,
since \\x — p(x)\\ — dist(x,C) is a Lipschitz function, we have that \\y\\ = 1,
\\yn\\ -+ 1. Because of the convexity of C, we have |(y + yn) G C and
thus \\\{y + yn)\\ > dist@,C) = 1. By the triangle inequality we have
Il2/n|| + ||j/|| > ||2/ + 2/n||,so ||^(j/ + 2/n)|| ~* 2. By the local uniform rotundity,
we get \\yn - y\\ -» 0.
8.38 Let C be a closed convex subset of a real Hilbert space H. Denote by
P the nearest point map of H onto C. Show that P is 1-Lipschitz, and if
we define f(x) — ^(\\x\\2 — \\x — P(x)\\2), then / is convex and f'(x) = P(x)
in the Frechet sense.
Hint: (x-P(x), z-P(x)) < 0 for all z G C. Indeed, if z G C and 0 < i < 1,
then zt = tz + A - t)P(x) G C and thus ||ar - P(x)\\ < \\x - zt\\ =
11 (a; — P(x)) — t(z — P(x))\\. Taking the square and expanding gives 0 <
-2t(x - P(x),z - P(x)) + t2\\z - P(x)\\2. Dividing by t and taking the
limit as t —> 0 establishes the claim. Thus, {x — P(x),P(x) — P(y)) >
0 and (P(y) — y,P(x) — P(y)) > 0. Adding these inequalities, we get
(x-yJP(x)-P(y)) > \\P(x)-P(y)\\2, which implies that Pis 1-Lipschitz.
Wehave2/(x) = ||^||2-inf{||x-y||2; y G C} = sup{2(z, y) - ||y||2; yG
C}, and thus / is convex as a supremum of affine functions. Fix x G H.

270 8. Differentiability of Norms
For every y G H, we have ||(;r + y) — P(x -f y)\\ < \\(x + y) — P(#)||, so
||(* + y) - P(x + y)\\2 < \\x + y\\2 - 2(x + y, P(x)) + \\P(x)\\2 = \\x + y\\2 +
H* - P(x)\\2 - ||*||2 - 2(y,P{x))- hence f(x + y) - f(x) - (P(x),y) > 0.
On the other hand, since ||* — P(*)|| < ||a; — P(x + y)\\, we get f(x + y) —
f{x) - (P(x), y) < (y, P(x + y) - P(x)) < \\y\\ • \\P(x + y)- P(x)\\ < \\y\\2,
which gives the differentiability assertion.
8.39 Let C = {x = (x7) G ^>(r); z7 > 0 for all 7 G T}; that is, C is
the positive cone in ^(L). Let P assign to x G ^(r) its nearest point
in C. Then P is a Lipschitz map (the previous exercise). Show that if T
is uncountable, then P is nowhere Gateaux differentiate. If T is infinite,
then P is nowhere Frechet differentiate. Note that Preiss proved that every
real-valued Lipschitz function on ^(r) is Frechet differentiable on a dense
set.
Hint: P(xO = x+ and (P(x)) = sign (x+). To see the Frechet case, if for
example x = (z7) and x1 > 0 for all 7, let a7 = — 2x1 for all 7 and it is
not true that a~1[P(x-\-a1e1) — P(x) — a7(P(#)) (e7)] —» 0. The Gateaux
case follows, because every x G ^(r) is countably supported and we use
the formula for P(x).
8.40 Let the norm || • || of a Banach space X be Gateaux differentiable. Let
P be a norm-one projection of X onto P(X) C X. Show that if x G Sp(x),
then ||s||'eP*(X*) = P(X)*.
Hint: \\x\\f and P*(||x||/) are two supporting functionals of Bx at x.
8.41 Show that the nonlinear operator <p from L2[0,1] into I/2[0,1]
defined by (p(x):t »—> sin(#(^)) is everywhere Gateaux but nowhere Frechet
differentiable.
Hint: (Lindenstrauss) (sin ?)'(/*) = (cosx)(h). <?>'(()): sin(aj) — x — (sin(l)—
l)z, which is not o(||x||) for x = X[o,*]-
8.42 Let X,Y be Banach spaces, and let T be an isometric mapping of
X onto Y. The Mazur-Ulam theorem claims that if T@) = 0, then T is
necessarily linear. Show this result for the case when Y is strictly convex.
Hint: Since T is continuous and T@) = 0, it is enough to show that
r(=±*) = \{T{x)+T{y)). Because ||2±* - *|| = ||=±* - y\\ = ||^||
and T is an isometry, we have
||r(?±*)-T(*)|| = \\T(!*L)-T(y)\\
= \r-^w=mT(*)-T(y))\i
If x,yeY, then{^} = {u€Y; ||u- x|| = ||u - y\\ = H^}. Indeed,
one inclusion is clear. Assume we have such a u. Then li"~g+y~"l| = llx~yN)
so u — x, y — u, and their center ^^- are all on the sphere of radius Hi^Ji;
hence, from the strict convexity of Y, we have that u — x = y — u.

8. Differentiability of Norms 271
Using this for u = T(^), T(x), and T(y), we then have T(*±*) =
L(T(x)+T(y)).
8.43 Show that the norm || • || on a Banach space X is LUR if and only if
lim||?n — ?|| = 0 whenever xn, x G Sx satisfy lim||xn -f x\\ — 2.
Hint: Using the proof of Fact 8.11, (ii) => (iii), show first that for xn,x as
in the definition of LUR, we have ||xn|| —> \\x\\.
8.44 Let xn G Sx and yn G Sx be such that lim \\xn + yn\\ — 2. Let zn
n—>-oo
be a point on the line segment between xn and yn for every n. Show that
lim ||z„|| = 1.
n—voo
Hint: Let fn G Sx* be such that /n(|(^n+2/n)) = |lkn+2/n|| —» 1- Observe
that f(xn) —> 1 and /(yn) —>¦ 1- Indeed, if f(xn) < 1 — 8 for all n, then
f(ym) > 1 + */2 for m large enough, contradicting ||/|| = 1, \\ym\\ = 1.
Then also /(zn) —> 1> and hence 1 > ||zn|| > fn(zn) —> 1.
8.45 ([MoTo]) Let X be a Banach space. Given x $ Bx> define R(x, Bx) —
conv({#} U Bx) \ Bx (cf.j Exercise 8.50). Prove that X is LUR if and only
if lim diam(R(tx,Bx)) = 0 for all x G Sx-
Hint: Given x G Sx and e G @,2], define 5(x,e) = inf{l-||^||; y G
Sx,||? — 2/|| > ^} (c/., Definition 9.1). Define A(x,e) = ^8(x,e); it is
a strictly increasing function of e. Then prove that diam(iZ(/x, Bx)) <
2[t(A(xr))~1Bt-^-) +t- 1]. This proves the only if part.
Now assume that X is not LUR. Find x G Sx, ? > 0, and a sequence
{zn} in S* such that \\xn - x\\ > e and ||?l^|| > 1 - ?, n G N. For a
given n — 3,4, 5,... take / G (^zj3 2). Then tXn?x is a convex combination
of xn and -^x. Evaluate the distance between tXn?x and 2~ix-
8.46 Assume that X is a reflexive Banach space whose norm is LUR (resp.
Frechet differentiable). Let Y be a closed subspace of X. Show that the
quotient norm of X/Y is LUR (resp. Frechet differentiable).
Hint: The LUR case: Let x,xn G Sx/y and \\x + xn\\ —> 2. Choose ?n G
A + e)Sx H ?n and x G Sx n ?. Then 2 + 2 > ||ar + xn\\ > \\x + xn\\ -> 2.
Use LUR to get \\xn — #|| —> 0. This implies ||? — ?n|| —» 0. The Frechet
case is similar, use Lemma 8.3.
8.47 Let (Xn, \\ • ||n), n G N, be Banach spaces. If their norms are LUR,
show that the canonical norm of {^2(Xn, || • ||n)J is LUR.
Hint: Let B||xn||2 + 2||x||2 - \\x + xn\\2) -> 0. Given e > 0, fix j0 such
OO OO
that Yl \\xj\\2 < e- By the convexity, it follows that also J2 lk?l|2 < ?
for large n. Then use LUR on the first jo coordinates.

272 8. Differentiability of Norms
8.48 Assume that the norm || • || of a Banach space X and its dual norm
on X* are both Frechet differentiable. Then the norm || • || and its dual
norm are both locally uniformly rotund.
Hint: Let x,xi,X2,... E Sx be such that \\x + xn\\ —> 2. Let fn be the
derivative of || • || at ^f^ and / be the derivative of || • || at x. Then ||/n|| —
1, fn(x + xn) = ||x + x„||, and ||/|| = 1 = f(x). If liminf (fn(x)) < 1, then
for some 0 < q < 1 and some subsequence n^, fnk(x) < q for all k. Then
fnk(x+xnk) < <l+fnk(xnk) < l+q < 2, a contradiction with fnk(x+xnk) =
\\x + Xnk\\ —> 2. Similarly, fn(xn) —> 1. Thus liminf(/n(ar)) = 1. Because
fn(x) < \\fn\\ \\x\\ - 1 for all n, lim(/„(x)) = 1.
Since || • || is Frechet differentiable at x, from the Smulian lemma we
have ||/n - f\\ - 0. Thus |(/n - /)(*„)| < ||/ - /n|||M| -> 0. Hence
f(xn) - fn(xn) + (/ - fn)(xn) -^ 1- Because the dual norm of || • || is
Frechet differentiable at / and f(x) — 1, from the Smulian lemma we have
\\x — xn\\ —> 0. This shows that || • || is locally uniformly rotund. Since the
dual norm is Frechet differentiable, X is reflexive and we can follow the first
part of this proof to show that the dual norm of || • || is locally uniformly
rotund.
8.49 A norm || • || of a Banach space X is said to have the BR)-properiy
if {xn} is a convergent sequence whenever \\xn + xm\\ —> 2.
Show that every space whose norm has the Bi2)-property is reflexive.
Note that every separable reflexive space has an equivalent norm with
the BiJ)-property ([OdSc]).
Hint: If / E Sx* and xn E Sx satisfy f(xn) —¦>• 1, then \\xn + xm\\ —> 2.
Use Corollary 3.56.
8.50 ([Mon2]) Let X be a Banach space. The drop defined by x E X \ Bx
is the set D(x, Bx) = conv({a:} UBx)- The Banach space X is said to have
the drop property if, given any closed set S C X such that S H Bx — 0,
there exists sG5 such that D(s, Bx) H S = {z}.
(i) Let X be a Banach space. Show that, given a closed set A C X such
that dist(A,Bx) > 0, there exists a E A such that D(a,Bx) C\ A — {a}
([Dan]).
(ii) Prove that X has the drop property if and only if X is reflexive and
has the Kadec property (that is, norm and weak convergent sequences in
Sx are the same).
Hint: (i): Use Theorem 3.53.
(ii): Assume that X has the drop property. If X is not reflexive, by
Corollary 3.56, there exists / E Sx* that does not attain its norm. Let
?n > 0, en —> 0. Choose xi € X such that f(x\) > 1, then &i E Bx such
that /(&i) = 1 — ?i. Choose x>i E [&i,?i] such that /(#2) = 1 + ?i, then
&2 E i?x such that /(&2) = 1 — ^2, and so on. The set {xn} is closed and does
not intersect Bx] however, xm E D(xn, Bx) for m > n, a contradiction.

8. Differentiability of Norms 273
If X does not have the Kadec property, we can find a sequence {xn} in Sx
and x E Sx such that xn ^ x but not xn —¦> #. Let ?n > 0, ?n —> 0. Define
zi — SqXi, choose z2 E [«i,^i], \\x\ - z2\\ < e2) then choose z3 E [#2,^2],
\\x2 — z31| < ?2, and so on- Apply the same argument as before.
Assume that X is reflexive and has the Kadec property. Let {xn} be a
sequence in X that is not eventually constant such that xn+\ E D(xnyBx)
for every n. Suppose first that ||xn|| /> 1. By (i), it must have a convergent
subsequence. If ||^n|| —* 1, by the Eberlein-Smulian theorem there exists
a subsequence that ^-converges to some xq. Because xo E conv{xn}, it
follows that I \xq 11 = 1. By the Kadec property, xn —> xq, and this implies
that X has the drop property.
8.51 Define a norm ||| • ||| on C[0,1] by ||x||2 = \\x\\l + \\x\\l where || • 1^,
resp. || • ||2, denote the norms of C[0,1] and I^fO, 1]3 respectively. Show that
I ¦ I is strictly convex (rotund) but not LUR on C[0,1].
Hint: To see that ||| • ||| is not LUR, consider x — 1 identically, and xn is a
function of the broken line through the points @,0), (±, 1), A,1). ||| • ||| is
strictly convex by Fact 8.11.
8.52 Define the following norm on t\\ ||ar|| = ||x||i -f ||ar||2. Then || • || is the
dual norm to some norm || • 1^ on cq.
Show that || • || is strictly convex and on its unit sphere the norm and
w*-topologies coincide. Consequently, || • \\^ is Frechet differentiate, yet its
dual norm || • || is not LUR.
Hint: To see that it is not LUR, use (±,0,0...) and @, ?,..., ?,0,...);
there are n coordinates equal to ±.
8.53 Let Y be a closed subspace of a Banach space X.
(i) Show that if the dual norm of X* is strictly convex, then every
continuous linear functional on Y can be uniquely extended to a functional on
X of the same norm.
(ii) Assume that the dual norm of X* is LUR. To every / E Sy* assign
as $(/) the unique extension of / on X. Show that $ is a continuous map
from Sy* into Sx*-
Hint: (i): The existence of such an extension is guaranteed by the Hahn-
Banach theorem. If F and G are two such extensions of /, consider the
norm of F + G in X* by calculating it on Y.
(ii): If fneSy*-^fe SY*, then ||$(/n) + $(/)|| > ||/n + /|| - 2.
8.54 Let {ez} be a Schauder basis of a Banach space X.
(i) Show that there is an equivalent locally uniformly rotund norm ||| • ||| on
X such that {e2} is monotone in ||| • |||.
(ii) Assume that X* is separable. Does there exist an equivalent Frechet
differentiable norm ||| • ||| on X such that {e2} is monotone in ||| • |||?

274 8. Differentiability of Norms
Hint: (i): First, renorm the space X by ||#||i = sup ^ az-ez- for x —
n<m"i=n "
^a2e2-. Then use the same method of proof as in Kadec's renorming
theorem; instead of the distances, we use the functions pn(x) = \\x — Pn(x)\\)
where Pn are the canonical projections for {ez}.
(ii): Not in general. Indeed, then {e2-} would be shrinking by
Proposition 8.34. However, James's space J has a separable dual and a
non-shrinking basis.
8.55 Let X be a Banach space. Show that if E} F are finite-dimensional
subspaces of X such that dim(F) > dim(i?) (no assumption on inclusion
of E and F)) then there is x G F with ||#|| — dist(;r, E) — 1.
Hint: First, assume that the norm of X is strictly convex (rotund). For
every x G F with ||a?|| = 1, let p(x) be the unique point in E for which
\\x — p(#)|| = dist(?,i?). Then p is continuous (see above and recall the
compactness of the ball in the finite-dimensional space). We have p(—x) —
—p(x). Since dim(F) > dim(E'), by the Borsuk antipodal theorem we get
x with \\x\\ — 1 in F such that p(x) — 0 (i.e., dist(#, E) — 1). The general
case follows by approximating the norm of X by strictly convex norms and
taking the limit of the points obtained.
8.56 Does there exist a bounded linear operator from C[0,1] onto H\1 Does
there exist a bounded linear operator from C[0,1] onto Co?
Hint: No for the first question. Otherwise, l^ is isomorphic to a subspace
of C[0,1]*, which is not the case. One of the reasons for this is that C[0,1]*
admits an LUR norm and l^ does not (see, e.g., [DGZ3] and [Trol]).
Therefore, C[0,1] does not have a complemented subspace isomorphic to
Yes for the second question (Sobczyk's theorem).
8.57 Let A be a subset of a metric space (X, p). We say that a point x G A
is diametral if sup{\\x — y\\) y G A} — diam(^4).
Define A = {x(t) G C[0,1]; 0 = x@) < x(t) < x(l) = 1}. Show that
every point of A is a diametral point.
Hint: Draw a picture.
8.58 Show that every separable Banach space X can be renormed by an
equivalent norm so that every convex set with more than one point has a
non-diametral point.
Hint: Let {/;} C Sx* be a separating set. Define ||| • ||| by ||ar|2 =
||x||2 + ^2~zfi(x)- Then find similarly as in Kadec's theorem that
whenever 2||arn|||2 + 2\jyn\l2 - fxn -f ynf -> 0 and {xn} is bounded, then
fi(%n — y-n) —> 0. Let C be convex, diam(C) = 1, and x,y G C. Let
||(ar + y) - xn% —> 1. Then, since \x - zn|||, \y - xn\ < 1, we have that

8. Differentiability of Norms 275
fi(x — xn — (y — xn)) = fi(x — y) —> 0 as n —> oo. Since {/;} is separating,
we have x — y.
8.59 A closed convex bounded set C in a Banach space X is said to have
normal structure if every closed convex subset K of C containing more
than one point has a non-diametral point in K. Show that every compact
convex set in a Banach space has normal structure.
Hint: Assume the contrary. Let d — diam(iv). Choose X\ G K. Find X2 G K
with \\x\ — #21| = d. Choose X3 G K such that fl371"^2 ~x%\ ~ ^> e^c- Then
±x-^LZfi\\<1-±\\Xn+1-Xi\\<d,
8=1 2=1
n
so J2(d ~~ lkn+i — ?»||) = 0- Since ||?n+i — X{\\ < d for every i, we get
i=i
||#n+i — Xi\\ — d. Hence {xn} has no convergent subsequence.
8.60 Find an example of a norm on a separable Banach space X that is
Frechet differentiate on a dense set and yet X* is nonseparable.
Hint: Let || • || be an equivalent LUR norm on cq whose dual norm is LUR.
Consider its dual norm || • ||* on X — l\. The set B — {\\x\\'\ x G SCo} is
a James boundary of (co, || • ||). By Theorem 3.46, convE) = Bix. We will
show that B — Si1.
To this end, let / G Six and F G Si* be such that F(f) = 1. The slices
given by F form a neighborhood base in the norm topology of Bix at /.
This is due to the LUR property of the dual norm. Indeed, if F(fn) -* 1,
then 2 > ||/ + /n||* >_F(f + fn) - 2. Therefore, ||/-/„||* - 0 by LUR of
|| • ||*, so if / G S^ \ B, we have sup(F) < 1 — 6 for some 8 > 0. Then also
b _
sup (F) < 1 — 6, a contradiction with convE) = B^. Therefore B = Six.
convB
If / G -B, then /(a?) = 1 for some x G 5Co. If xn G Sco are such that
/(#n) —> 1, then again we find that 2 > ||a? + xn\\ > f(x + arn) —> 2 and
thus by LUR of || • || we have || X Xfi II = 0. By Smulian's lemma, / is a
point of Frechet smoothness, so || • ||* is Frechet differentiable on a dense
set in ?1 and t\ is not separable.
8.61 ([Sin2]) Show that if X is a separable Banach space such that its
second dual norm is Gateaux differentiable, then X* is separable.
Hint: We must show that X* is ^-separable (Proposition 3.26). It suffices
to show that the duality map x G Sx »—> || ¦ || (#) G Sx* is norm to weak
continuous.
Let xn,x G Sx and xn —> x. Let fn,f G Sx* be such that fn(xn) — 1
and /(ar) = 1. Then fn(x) = fn(xn) + fn(x - xn) —»¦ 1- Since a: G Sx~ is
a point of Gateaux differentiability of the second dual norm, by Smulian's
lemma we have fn ^ f in X***. Since fn, / G X*, we get /n -^> / in X*.
a?i + ...-+- xn
Vn + l

276 8. Differentiability of Norms
8.62 (Giles, Kadec, Phelps) Show that if X is a Banach space the third
dual norm of which is Gateaux differentiate, then X is reflexive.
Hint: By the Bishop-Phelps theorem, it is enough to show that every F ?
Sx** that attains its norm on Sx* is from X. Let F ? Sx** and / ? Sx*
be such that F(f) = 1. Let xn ? Sx be such that f(xn) —> 1. Consider
/ as an element of Sx*** and F ? Sx**** ¦ Since the third dual norm is
Gateaux differentiable at /, we get from Smulian's lemma that xn —> F in
X****. Since sn, F ? X**, we get that zn ^> F in X**. Since xn ? X and
X is w-closed in X**, we have that F ? X.
8.63 Let Y be a closed subspace of a Banach space X and x ? X. We say
that z is orthogonal to Y if ||ar + y|| > ||z|| for every y ? Y. Note that, in a
Hilbert space, this is equivalent to x _L Y (Lemma 7.51).
Prove the following. Let (X, || • ||) be a Banach space such that the dual
norm || • ||* is Frechet differentiable. Let Z be a subspace of X of finite
codimension. Let M be the set of all elements in Sx that are orthogonal
to Z. Then M is a compact set.
Hint: Note that (X/Z)* = ZL and M = {x ? SX) \\&\\x/z = Ikll), so by
the Hahn-Banach theorem, M is formed exactly by those elements x ? Sx
for which there exists / ? Sx* H ZL such that f(x) = 1.
Given / ? S^flZ1, there is a: ? Sx such that /(#) = 1 (find appropriate
x in X/Z, which is finite-dimensional) and it is unique, since || • ||* is Frechet
differentiable. Call x — <p(f). By the Smulian lemma, <p\ Sx* H ZL —> Sx
is norm-norm continuous and M — <p(Sx* fl Z1), so M is compact.
8.64 Let X be a separable Banach space whose norm is Frechet
differentiable. Show that if Y C X* is a closed 1-norming subspace of X*, then
Y = X*.
Hint: Given x ? Sx, put z = Hz]!'. Let {/n} ? 5y be such that /n(#) —» 1.
By Smulian's lemma, fn —> z, and hence z ? Y. Thus, Y contains a James
boundary {||ar||;; x ? Sx} of X. By Theorem 3.46, Y = X*.
8.65 Show that every exposed point of a convex set is extreme and give an
example of an extreme point that is not exposed.
Hint: Consider the function / that is identically zero on [—oo, 0] and f(x) —
x2 for x > 0. Check the origin and the epigraph of /.
This example also shows that an extreme point cannot in general be
replaced by an exposed point in Lemma 3.38.
8.66 Show that none of the spaces C[0,1], Co, or Za[0,1] is isomorphic to
a dual space.
Hint: The unit balls of these spaces are not the closed convex hulls of their
extreme points. Then use Theorem 8.29.

8. Differentiability of Norms 277
8.67 Let X be a Banach space with a strictly convex (rotund) norm. Show
that all points of Sx are exposed points of Bx •
Hint: If f(x) - f(y) = 1 for x ^ y G Sx, then /(|(x + y)) = 1 and thus
||?±*|| = 1.
8.68 Let C ^ 0 be a convex compact set in a Hilbert space H. Show that
C has at least one exposed point.
Hint: Assume that ||e|| — sup{||y||; y G C}. Then x is an exposed point of
||a:||BjHr (previous exercise) and thus an exposed point of C.
8.69 Show that exposed points need not form a James boundary.
Hint: For a two-dimensional example, consider the ball in R2 that is the
intersection of two disks of radius 2 centered at the points A,0) and (—1,0)
and draw a picture of the dual ball.
8.70 Let X be a Banach space. Show that if the norm of X is LUR (in
particular, if X is a Hilbert space), then every x G Sx is a strongly exposed
point of Bx-
Hint: If x G Bx, / G Bx* satisfy f(x) = 1, and xn G Bx are such that
f(%n) —* 1, then 2 > \\x + xn\\ > f(x + xn) —> 2. Hence \\x + xn\\ —> 2 and
thus ||x —a?n|| —> 0 by the local uniform rotundity or the parallelogram law.
8.71 Let K = conv{{0} U {en}?°=1} C i2, where en are the standard unit
vectors. Show that there does not exist a probability measure // on K
supported by {en} that represents 0. This shows that Choquet's representation
theorem cannot give a measure supported by the strongly exposed points
{en}ofK.
Hint: Check that {en} is the set of all strongly exposed points of K. Choose
8.72 Show that every standard unit vector ez- is a strongly exposed point
ofS^.
Hint: The unit vectors e2- are points of Frechet differentiability of the norm
of Co- Then use Smulian's lemma.
8.73 Let C = {x G Bi2] X{ > 0 for all i). Show that 0 is an exposed point
of C but a not strongly exposed point of C.
If C is defined in an analogous way for a nonseparable ^(r), then 0 is
not even exposed. Is it extreme?
Hint: Exposed: Consider the functional B~l). Strongly exposed: en -^ 0,
so f(en) —*• /@), yet en ¦/-> 0. In the nonseparable case, every continuous
linear functional has a countable support and thus must vanish at some
unit vectors. Hence 0 cannot be exposed.

278 8. Differentiability of Norms
However, 0 is extreme, as we check by coordinates. Note that 0 is then
not even a weak G^-point in C (i.e., an intersection of a countable family
of weakly open sets) for a similar reason as for non-exposedness.
8.74 Let C be a compact convex set in a Banach space X and x be an
exposed point of C. Show that x is a strongly exposed point of C.
Hint: Use the definition and compactness.
8.75 Let C ~ {x G BiY\ X{ > 0}. Show that C is not a closed convex hull
of its u?*-strongly exposed points. A strongly exposed point of C C X* is
called w* -strongly exposed if it is strongly exposed by a functional from X.
Hint: The only t*;*-strongly exposed points are the standard unit vectors.
0 is not in their norm closed convex hull because it is separated from them
by the functional A,1,...). This functional strongly exposes 0 in C.
8.76 Prove that every extreme point of Bt^ is a w*-exposed point of Bt^.
Hint: Ext(Bioo) = {(ar,-); x{ - ±1}; use y G Btl with ?)yt-sign(a;j) = 1.
8.77 (Lindenstrauss, Phelps [LiPh]) Show that if X is a separable reflexive
Banach space, then there is an equivalent norm || • || on X such that the
unit ball in || • || has only countably many strongly exposed points.
Hint: Let TV be a maximal ^-separated set in Sx- By contradiction, prove
that C = conv(iV) contains \Bx- let x G \Bx \ C. By the separation
theorem, there is / G Sx* such that f(x) > swpN(f). For any 6 > 0, there
is y G Sx such that f(y) > 1 — 8. By the maximality of N, there is z ? N
with § > ||j,-z|| > f(y)-f(z). Thus supw(/) > f(z) > /(y)- | > 1-8- \.
Hence l-\-8< supAr(/) < f(x) < \\x\\ <l-|,sol-|< supN(f) <
Put \\x\\ = fic(x)- Since X is reflexive, C is the closed convex hull of its
strongly exposed points, so the set of the strongly exposed points is infinite.
Because iV is in a separable space, we have card(JV) < Ko, so we just show
that every strongly exposed point of C lies in N.
Let x G C be a strongly exposed point of C. Then there is / G X* such
that f(x) — supc(/) and whenever f(xn) —> /(#) for some #n G C then
zn —> x. Note that supc(/) — supiV(/), and thus there are yn E N such
that f(yn) —> /(^)- Thus yn —+ x. Because Af is closed (it is ^-separated),
we get x G N.
8.78 (Lindenstrauss, Phelps) Show that if X is an infinite-dimensional
reflexive Banach space, then Ext(Bx) is uncountable.
Hint: Assume Ext(Bx) = {xn}- Put Fn = {/ G J5X*; /(*„) = ||/||}. Show
that each Fn is weakly closed. By Lemma 3.38, we get that Bx* is the
union of {Fn}. By the Baire category theorem, one of Fn, say i<\, has a
relative interior point /o in Bx* in its w-topology. Assume ||/0|| < 1. Thus,

8. Differentiability of Norms 279
there are t/i,... ,2/n ? X and e > 0 such that / G Fi whenever ||/|| < 1
and |(/ — fo)(yi)\ < ? for all i = 1,. .., n. Let
N = {/ G X*; /(y,-) = /<,(&¦) for i = 1,..., n and /(xx) = /0(*i)}.
Because X is infinite-dimensional, there is g G AT D Sx* • Since g G N, we
have g E F\ and thus 1 = ||#|| = #(xi) = /o(#i) = ||/o||, a contradiction.
8.79 Let X be a separable reflexive Banach space. Show that extreme
points of Bx are not all isolated in the norm topology.
Hint: Every set of isolated points in a separable metric space is countable.
Then use the previous exercise.
8.80 Let C be a closed convex bounded set in a Banach space X. Show
that C is weakly compact if and only if C and C Cl** have the same
extreme points.
Hint: If C is not weakly compact, by Theorem 3.55, let / G X* not attain
its supremum on C. By the Krein-Milman theorem, there is an extreme
point x of C such that f(x) — supc(/) and so x (? C.
8.81 ([Monl]) We say that a subset A of a Banach space X has property (*)
if A is a nonempty, closed, convex, and bounded subset of X and every point
a G A is a proper support point] that is, given a G A, there exists a* in X*
such that a* (a) = supyl(a*) and there is x G A such that a*(x) < supA(a*).
Show that:
(i) C[0,1]* contains a subset with property (*).
(ii) Given an uncountable compact set K, C(K)* contains a subset with
property (*).
(iii) If T is an infinite set, ^oo(r) contains a subset with property (*).
(iv) If X contains an isomorphic copy of -^i(r), then X* contains a subset
with property (*).
Hint: (i): Define A = {/i G C[0,1]*; \\fi\\ = 1,// > 0,/i atomic}. A is a
convex and bounded set. Prove that A is also closed. To check that every
fiQ G A is a proper support point, find a countable set D C [0,1] such that
A'oQO, 1] \ D) — 0. Define a continuous linear functional L on C[0,1]* by
L(fi) = fJ.(D), fi G C[0,1]*. Then L supports A properly at /x0.
(ii): Use Milyutin's theorem: All C(K) spaces are isomorphic for K an
uncountable compact metric space.
(iii): Assume first that V is countable. C[0,1] is then isometric to a
quotient of ^i(r). Then ^oo(r) has a closed subspace isometric to C[0,1]* and
the result follows from (i). If V is uncountable, i^ is isometric to a closed
subspace of ^oo(r), and the result follows from the first part.
(iv): X* then has a quotient that is isomorphic to ^oo(r). Let q be the
quotient map. Using the previous results, find a subset Aq with property
(*) in this quotient and define A — q~1(Aq) C\ (M + e)Bx*, where e > 0 is
arbitrary and M is a bound for Aq in the norm.

280 8. Differentiability of Norms
8.82 Let X be a reflexive Banach space whose norm is Frechet differen-
tiable. Show that if A\, A2 are bounded closed convex subsets of X such
that A\ H A2 — 0, then there are balls B\,B2 such that A\ C B\, A2 C B2)
and Bi n B2 = 0.
Hint: Proof of Theorem 8.33 and Theorem 3.17.
8.83 Let X be a Banach space. We say that X has the Kadec-Klee property
if the weak and norm topologies coincide on Sx •
We say that X* has the w*-Kadec-Klee property if the w*- and norm
topologies coincide on Sx* •
(i) Let X be a locally uniformly rotund space. Show that X has the Kadec-
Klee property.
(ii) Let X be a Banach space such that the dual norm of X* is LUR. Show
that X* has the w*-Kadec-Klee property.
Hint: (i): If xn -^ x in Sxy take / G Sx* such that f(x) — 1. Then
/fan) —> 1 and ||s + xn|| > f(x + xn) —> 2. By LUR, xn —> x.
(ii): Assume that /n, / G Sx* satisfy /n ^ /. Given ? > 0, get x G Sx
such that f(x) > 1 - ?. Then ||/n + /|| > (fn + /)(«) > 2 - ? for n large
enough. Thus ||/n + /|| —» 2, and use LUR.
8.84 Let X be an infinite-dimensional Banach space. Show that the weak
and norm topologies do not coincide on Bx •
Hint: 0 is in the weak closure of Sx (Exercise 3.8).
8.85 Show that no point x G BCo has the property that the norm and the
weak topologies of BCo coincide at x.
Hint: Given x G BCo and /i,...,/n GcJ, use the fact that each /; is given by
a summable sequence to find a point x' G BCQ such that max \fi(x — x')\ < e
and \\x — x'Hoo is larger than |.
8.86 Let X be a Banach space. Let || • || be an equivalent norm on X* such
that the w*- and norm topologies coincide. Show that then || • || is a dual
norm on X*.
Hint: Assume, by contradiction, that a net {f\} in Sx* converges to / in
the w*-topology and ||/|| — 1 + ? for some e > 0. Let f\ G A + e)Sx* H
{/ + t(f\ — /); t > 0}. Since \\f\ — f\\ > ?, the numbers t\ that define f\
are bounded and thus we have f\—>f because f\, f G A -f s)Sx* • Thus
fx —> /, which is a contradiction since ||/a — /|| > ||/a — /|| > ^ (draw a
picture in two dimensions).
8.87 Let X be a separable Banach space. Show that if X* has the w*-
Kadec-Klee property, then X* is separable.
Hint: Let xn be dense in Sx and fn G S^ be such that fn(xn) = 1.
Since {/n} is separating, conv^ {/n} = Bx*, so, given # G 5x*, there are

8. Differentiability of Norms 281
w*
gn E conv{/n} such that gn —> g. Then gn —> g by the iu*-Kadec-Klee
property. Since conv{/n} is norm separable, the statement follows.
8.88 Let X be a separable Banach space and assume that X* has the w*-
Kadec-Klee property. Show that if C is a u>*-compact convex set in X*,
then C has a point where the identity map of C into C is w*-to-norm
continuous.
Hint: Consider the dual norm || • || on C. Since it is lower semicontinuous
on (C, iu*), which is a complete metrizable space, by the Baire category
theorem it is continuous at some point of C. At this point, due to the
i*;*-Kadec-Klee property, the w*- and norm topologies coincide.
8.89 Let X be a Banach space. Show that if the dual norm of X* is LUR,
then every point of Sx* has a neighborhood base of the relative norm
topology of Bx* formed by slices given by functionals from X.
Hint: LUR implies tu*-Kadec-Klee. The w*-topology has a neighborhood
base formed by slices given by elements of X by Choquet's lemma.
8.90 (Troyanski) Show that if a Banach space X admits a strictly convex
(rotund) norm and also a norm with the Kadec-Klee property, then X
admits an LUR norm.
Note that there are spaces that admit Kadec-Klee norms and no norm
that is LUR (Haydon; see, e.g., [DGZ3]). The space i^ admits a strictly
convex norm but no LUR norm (Lindenstrauss, Troyanski; see, e.g.,
[DGZ3]).
Hint: (Raja [Raj2]) First, let || • || be a norm of X that is both strictly
convex and Kadec-Klee (Asplund averaging; see, e.g., [DGZ3]). Then one
can use Exercise 3.87 to see that each point of the new unit sphere is
denting; that is, for every x E Sx and e > 0 there is a halfspace H such
that x E H and diam(i?x OH) <e. For m E N, put
Am = {xG Bx] diam(i?x fl H) > -^ for all halfspaces H containing x}.
Check that Am is closed, convex, symmetric, and that 0 E Int(Am).
Let fm be the Minkowski functional of the set Am, which is an equivalent
norm on X. Let am > 0 be such that am/^(a:) < 2~m||a:||2 for all x ? X.
Define \x\2 = J2 amfm(x) + Ikll2- Then | • | is an equivalent norm on X.
In fact, | • | is LUR: let #, x^ E X be such that
2|af+ 2M2-|z +^|2^0. (*)
We must show that \x — Xk\ —> 0.
For x = 0 it is clear. Assume x ^ 0. Multiplying x and Xk by tAt, we
assume that ||z|| = 1. From (*) we get \\xk\\ —+ \\x\\ = 1, \\x + Xk\\/2 —> 1.
Replacing xk by x'k = xk/\\xk\\, we get ||sj.|| - 1 and 1 > \\x + x'h\\/2 ->
1. Fix any e > 0. Find m E N with m > 2. Since x, x u are denting,
x,x'k ? Am. Because Am is closed, fm(x) > 1 and fm(x'k) > 1- By (*).

282 8. Differentiability of Norms
/m((^ + xk)/2) > 1 f°r ^arge k E N. Hence for these k we have (x +
tfjj.)/2 ^ Am, and thus there is a halfspace H such that (x + x'k)/2 E # and
diam(Bx D H) < — < e/2. Since a?, xj. E 2?x, by a convexity argument we
have \\x — a?j.||/2 < e/2; that is, \\x — x'k\\ < e. This means that x'k —> x and
thus xjfc ¦—> x.
8.91 Assume that a Banach space X has the Kadec-Klee property. Show
that the norm and weak Borel structures on X coincide and X is a Borel
set in (X**,w*).
Note that the weak and norm Borel structures do not coincide on i^
(Talagrand).
Hint: (Schachermayer) The map (tyx) »—> tx is a Borel homeomorphism
@,oo) x (Sx,w) —* (X \ {0},u?) (that is, the mapping and its inverse
map Borel sets onto Borel sets). This follows from the fact that the first
coordinate of the inverse y h-> (||2/||,2//||j/||) is w-lower semicontinuous and
hence Borel, whereas the second coordinate is the product of the w-upper
semicontinuous function y i-> ipr and the continuous identity map.
Moreover, since (Sx,w) is completely metrizable by Kadec-Klee, it
is a Gs set in its w*-compactification. By Goldstine's theorem, this
compactification is Bx** • Thus X is a Borel set in (X**, w*).
8.92 Let || • || be an LUR norm on a separable Banach space X. Show that
the set S of all elements in X* that attain their norm is dense and Gs in
X\
Hint: By the Bishop-Phelps theorem, S is dense in X*.
If / attains its norm at x and xn E Sx satisfy ||/|| = 1 = lim (f(xn)),
n—»-oo
then 2 > \\x -f xn\\ > f(x -f xn) —> 2 and, by LUR, \\xn — x\\ —> 0. Thus,
the dual norm is Frechet differentiable at /. Hence S coincides with the set
D of all points of Frechet differentiability (the other direction follows from
Exercise 8.19). The set D is a Gs set in X*.
8.93 Let T:$,2 —* f-2 be a diagonal operator defined for x — (#«•) E ^2 by
T(ar) = ((l — j)ff2-). Show that T does not attain its norm.
OO .y
Hint: Clearly, ||T|| = 1. If x E J3/2, then ? (l - i) x? < ?x? < 1-
2 = 1
8.94 (Lindenstrauss). Let Y be the space c$ in the standard sup-norm || • ||.
Consider an equivalent strictly convex norm ||| • ||| on c0 (it is a separable
space), and let Z = (c0, ||| • |||). Set X = Y 0 Z with the norm \\(y,z)\\ -
max{||t/||, HI}. Show that the set of all bounded linear operators from Z
into Z that attain their norm is not dense in B(X).
Hint: Let To be an isomorphism of Y onto Z with ||To|| = 1 and define
T E B(X) by T(y,z) = @,To(t/)). Denote e-^r < \ and assume that
there is f E B(X) with ||f-T|| < e and ||f || = \\T(y0,z0)\\ for some (t/0, *o)

8. Differentiability of Norms 283
in Bx. Put (u,v) = T(yo,2o)- Then \\u\\ < e and, since \\T\\ > e: we have
11w11 < ||T|| = \\v||. Since Sy has no extreme points, there is yi E Y \ {0}
such that 112/1 H-2/o11 - ||-yi+yo|| < 1- Hence \\f(yo,zo)±f{y1,0)\\ < ||f ||.
Since Z is strictly convex and ||v|| = ||T||, we have T(yi,0) — B/2,0) for
some y2 G Y. Then 6:||s/i|| > ||T(yi, 0) - f (yij 0)|| > ||T0(yi)|| > 2e\\yi\\, a
contradiction.
8.95 Let XjY be isomorphic Banach spaces. Denote by d(X,Y) the
Banach-Mazur distance between X and Y,
d(X, Y) = inflllTIIHT-1!!; T an isomorphism of X onto Y}.
Note that d(X,Y) > 1 and we have d(X,Z) < d(X, Y) d(Y, Z). The fact
that d(-^, Y) < d means that there is an isomorphism T of X onto Y such
that —By C T(BX) C C2#y for some ci and C2 satisfying c\c^ < d.
Show that if X is an n-dimensional Banach space, then d{X)l7l) < n.
Therefore, if X and Y are n-dimensional spaces, then d(^, Y) < n2.
Hint: Let {#;;/;} be an Auerbach basis of X. Let X\ be the space X
renormed by the norm whose unit ball is given by conv{x2}, and let Xqq
be the space X with the unit ball determined by the parallelepiped on
{xi}. Then BXl C Bx C #xTO. Check that BXoo C nBXl. Therefore
BXl C Bx C n?Xl.
8.96 Let X and Y be n-dimensional spaces such that d(X, Y) = 1. Show
that X and Y are isometric.
Hint: Use a compactness argument. If T is an isomorphism of X onto Y
with IITIHIT-1!! = 1, then the map f = HT)^ is an isomorphism of X
onto Y that satisfies ||T|| = H^-1!! = 1, so T is an isometry.
8.97 Find an example of two Banach spaces whose Banach-Mazur distance
is 1 but these spaces are not isometric.
Hint: (see [PeBe]) For X, take cq with the norm
00 i_
||z||i = fmax|^|2 + ^2_i|x2-|2J2,
2 = 1
and for Y take Co with the norm
00 i_
|H|2= (max|^|2 + ^2-i+1|^|2)'.
2 = 2
Define Tn ? B(X,Y) by Tn(x) - (xn,xi}.. .,arn_i,a?n+i,...). It is clearly
onto, ||Tn|| -> 1, and WT^W -> 1. Therefore d(X,Y) = 1. As in
Theorem 8.13, we find that X is strictly convex. However, Y is not strictly
convex (consider (—1,1,0,0,...) and A,1,0,0,...)). Therefore, X and Y
cannot be isometric.

284 8. Differentiability of Norms
8.98 Numbers Ci, Ci in the definition of equivalence of basic sequences
may serve as a measure of "closedness" of basic sequences.
Show that if basic sequences {e2}, {/;} are equivalent with constants Ci,
C2, then the spaces span{e2}, span{/2} are isomorphic and we can estimate
d(span"{ef}5span{/f}) < CiC2.

9
Uniform Convexity
Definition 9.1
Let (X, || • ||) be a Banach space. For every e G @, 2], we define the modulus
of convexity (or rotundity) o/|| • || by
8x{e) = inf{i- |^y^|; *,ye bx,\\*-y\\ > *}¦
The norm || • || is called uniformly convex (UC) (or uniformly rotund (UR))
if fix (z) > 0 for all e G @,2]. The space (X, || • ||) is then called a uniformly
convex space.
Note that 6x(z) — inf{5yF:); Y is a 2-dimensional subspace of X}.
Lemma 9.2
Let (X, || • ||) be a Banach space, and let 6(e) be the modulus of convexity
of\\.\\. Then %) = inf{l-||^||; x,yeSx,\\x-y\\ = e}.
Proof: (see, e.g., [Fig]) First, note that if x,y G Bx and \\x — y\\ > e,
then there are x1\yf on the segment [x}y] such that x ^ — ^^- and
\\x' — j/W = e. Thus it is enough to consider x, y G Bx with \\x — y\\ — e. It
remains to prove that
sup{||z+y||; x}yeBx,\\x-y\\ = e) = sup{||*+2/||; *> 2/ G SXl \\x-y\\ = e}.
We may assume that X is two-dimensional, so that the suprema are
attained. Assume that uq,vq G Bx maximize the left-hand-side supremum.
We will show that uo,vo € Sx- By contradiction, assume that ||vo|| < 1.
Denote A = {w G #x;||w — uo\\ = ?}• Let x* G Sx* be a functional
such that x*(uq + t>o) = ||^o + ^o||. Then, for w G A, we have z*(uo + w) <

286 9. Uniform Convexity
Hwo + HI ^ \\uo + vo\\ — x*(uq + vq). Since ||vo|| < 1, it follows that x*
attains a local maximum with respect to A at vq. This implies that x*
norms (vq — uq), so x*(vq — uq) = \\vq — uq\\ = e. Then we get ||i*o|| <
§(lK + M\ + \\vo - Uq\\) = %(x*(u0 + Vq) + X*(v0 - Uq)) - X*(v0) < 1.
This is not possible. Indeed, take 6 — |min(l — ||i«o||> 1 — ||^o||) > 0, then
v! = uq + <$(uo + vo) G5jf,i;/ = r;o + 6(i«o + vo) ? #x,
?, and
lit' -f r/|| = A + 25)||wq + fo|| > ||^o + ^o||, contradicting the maximality.
D
Note that if ||#|| =
\x + y
— 1 and \\x — y\\ = ?, then
x +
y-x
>\\x\
y-x
> 1
and thus we have 6(e) < § for every e E [0,2].
It is easy to see that l\ and l^ are not uniformly convex. Consequently,
Co, ?i> and t^ are not uniformly convex. On the other hand, if H is a
Hilbert space and || • || is the Hilbertian norm of H, then || • || is uniformly
convex.
Indeed, using the parallelogram equality, we have for e E @, 2],
{II x ~\~ v II 1
i — II —^— jj; x,yeSx,\\x-y\\ = ej
= ^-S
+ il
x-y
; xyyeSx,\\x-y\\ = ej
= 1
^i
>0.
Theorem 9.3 (Clarkson; see, e.g., [Disl])
Let (fi, fi) be a measure space. Ifp E A, co); then Lp(fj) is uniformly convex.
In the proof, we will use the following one-dimensional fact.
Fact 9.4
Let p E A, oo) and e > 0. There is 8(e) > 0 such that if numbers x,j/GR
.^e))(l?l!+Iidl).
satisfy \x — y\ > ?max{|#|, \y\}, then |^|^-|P < A
Proof: By homogeneity, we may assume that x — 1 and 1 — e > y >
0. We have [(^)P]' = .KW > flT1 = [^f for 2/ € @,1).
Consequently, /(y) = ^ (^")P is a decreasing function on [0,1].
Thus f(y) > /(l - e) > f(l) = 0 for y E @,1 - ?), and the existence of
6(e) follows.
D
Proof of Theorem 9.3: Let e E @,2] be given and let 6 = 8(e-4~i) be
from Fact 9.4. Let #, y E Lp(fx), \\x\\, \\y\\ < 1, and ||x- — y|| > ?, where || • ||
is the canonical norm of Lp(jjl). Put
M = {W; e?(\x(u)\* + \y(w)\") < 4|x(w) - y(w)|p}.

9. Uniform Convexity
287
+1
We claim that max{JM \x\p dfi, fM \y\p dfi) > —
Assuming this claim is true, we can finish the proof as follows. Using the
convexity of the function \x\p, Fact 9.4, and the claim, we have
\x(u)\r + \y(u>)\*> \x(u>) + y{u)\
p\
) dfj,
Therefore,
/
x(u) + y(u) \p
>
>
\x(u)\r+\y(u)\P \x(u>) + y{u)
2 I 2
¦x(u>)\P + \y(u>)\P}dti> Sep
~ 21
Jdfj,
')dfi> -TTT-
dfx <
i
\x(oj)\p + \y(w)\P
dfi— 6 —
2p
+2
< i-s—
±+2
Hence |||a: + y||< (l - 6-^) \ so 6(e) > 1 - (l - 6-j^) P > 0.
To prove the claim, consider the complement Mc of the set M. We have
/ IzH-yHI'd/x < ?-\ (|x(W)|" + |y(W)|")dAi
< ejJ(\x(u)\r + \y(u)\r)dii<
e?
Therefore, using ||a; — y\\ > e, we have JM \x(u) — y(w)|p d(i > ^-. Hence
H^ — y\\M > ?2~i", where || • ||M denotes the norm of Lp(M,n). Thus
1 ?
max{||x||M,||2/||M} > - • —.
^ 2?
D
Fact 9.5
Let (X, || • ||) be a Banach space. The following are equivalent:
(i) X is uniformly convex.
(U) Ifxn,yn EX, nEN; lim B||xn||2 + 2\\yn\\2 - \\xn + yn||2) = 0, and
n—+oo
{#n} is bounded, then lim ||a:n — yn|| — 0.
n—>oo
(m) If xn,yn e Bx, n eN, and lim ||a?n + yn|| = 2, then \\xn - yn\\ - 0.
n—»-oo
Proof: Only (iii) => (ii) needs more work. Let {xn},{yn} C X, {xn}
bounded, and lim B||a:n||2 + 2||2/n||2 — \\xn -f yn\\2) = 0. From the estimate
n—>-oo
|2 . OIL. I|2
2||xn||2 + 2||2/n|
kn+yn|r > 2lkri|r+2||2/n||2-(||xn|| + ||yn|
= (lkn||-||2/n||J>0,

288 9. Uniform Convexity
it follows that lim (||^n|| — \\yn\\) = 0 and thus {yn} is bounded. By passing
n—KX)
to a subsequence, we may assume that lim \\xn\\ — lim ||yn|| = a. If a = 0,
n—»-oo n—KX)
we are done.
Assume a > 0, then ||arn + yn\\ —> 2a. We have Tifjr, yr^^2^!- ? i?x, and
?
Example
Define an equivalent norm | • ||| on C[0,1] by ||/||2 = ||/||^ + \\f\\l where
|| • H^ denotes the standard supremum norm of C[0,1], and || • ||2 denotes
the canonical norm of Z<2[0,1]. It was shown in Exercise 8.51 that ||| • ||| is
a strictly convex norm on C[0,1]. Consider functions fn = 1 and gn for
every n, where gn is the broken line determined by the points @,1), (^, l),
A,1). It is easy to verify that fn,gn fail the property of uniform convexity.
Thus, I • | is not uniformly convex.
Definition 9.6
Let (X, || • ||) be a Banach space. For r > 0, we define the modulus of
smoothness of\\ • ||;
r||z +r/i|| + ||z-r/i||-2 ^
p(r) = supj^ ! ; ||x|| = ||A|| = 1 j.
We say that \\ • \\ is uniformly smooth if lim ^-^ — 0. We then say that
(X, || • ||) 25 uniformly smooth.
Note that 2||x|| = ||(x + rh) + (x - rh)\\ < \\x + rh\\ + \\x - rh\\, so p is
a nonnegative function.
A norm || • || is uniformly smooth if for every e > 0 there is S > 0 such
that for all x ? Sx and y ? 6Bx we have \\x + y\\ + \\x — y\\ < 2 + s\\y\\-
Clearly, a subspace of a uniformly smooth space is uniformly smooth.
Fact 9.7
Let (X, || • ||) be a Banach space. The following are equivalent:
(i) X is uniformly smooth.
\\x + th\\ - \\x\\
(ii) The limit lim ¦ — = ||?||'(/i) ? X* exists uniformly for
x,h ? Sx •
(Hi) The norm is Frechet differentiate on Sx and the map x i—> WxW from
Sx into Sx is uniformly continuous.
We omit the proof of Fact 9.7 since it is similar to that of Lemma 8.3.
Definition 9.8
Let f be a real-valued function on an open subset U of a Banach space X.

9. Uniform Convexity 289
We say ihai f is uniformly Gateaux differentiable (UG) on U if for every
f(x + ih) — f(x)
h G Sx ihe limii lim = f(x)(h) G X* is uniform in
t->o t
xeu.
We say thai f is uniformly Frechet differentiable (UF) on U if the limii is
uniform in x G U and h G Sx-
Note that / is uniformly Frechet differentiable on an open convex set U
if and only if it is Frechet differentiable at every point of U and the map
x »—> f'{x) is uniformly continuous as a map U —> X*.
The norm || • || of a Banach space X is called uniformly Frechet
differentiable (or UF-smooth), resp. uniformly Gateaux differentiable (or
UG-smooth), if || • || is uniformly Frechet (resp. uniformly Gateaux)
differentiable on Sx •
Lemma 9.9 (Lindenstrauss; see, e.g., [LiT3])
Lei (X, || • ||) be a Banach space, and let 8(e) be the modulus of convexity of
|| • || and p*(r) be the modulus of smoothness of the dual norm \\ • ||*. Then,
for every r > 0,
p*(r) = sup{r|-«(e); 0 < e < 2}.
Similarly, let p(r) be ihe modulus of smoothness of \\ • || and 8*(e) be the
modulus of convexity of the dual norm || • ||*. Then, for every r > 0,
p(r) = sup{r|-6*(^); 0 < e < 2}.
Proof: We claim that for e G @, 2] and r > 0 we have 6(e) + p*(r) > r|.
Indeed, let x,y G Sx be such that \\x — y\\ > e. Choose /, g G Sx* such
that f(x + y) — ||# + 2/|| and g(x — y) — \\x — y\\. By the definition of p*(r),
we have
2p*(r) > ||/ + ^|r + ||/-r^|r-2
> (f + Tg)(x) + (f-Tg)(y)-2
- f(x + y) + rg(x - y) - 2
= Ik + y\\ + T\\x — s/|| — 2 > ||* + 2/|| + re-2.
Hence 2 — \\x -f y\\ > re — 2p*(r). Thus, from the definition of 8(e) we have
8(e) + p*(r) > r§. Consequently, p*(r) > sup{r| - 8(e)] 0 < e < 2}.
To prove the converse inequality, let r > 0 and /, g G Sx* • For rj > 0,
there exist x,y G Sx such that (f+rg)(x) > \\f+Tg\\*— r] and (f—rg)(y) >
||/-rflf||* -77. Hence
< J
* A
< -
(/(*+
1 2
5A1*-
y) - 2) + ^(ar -») + »/
-l) + |||x-j/|| + ^
y||) + ilk-y|| + »?

290 9. Uniform Convexity
< sup{*§ - 6(e); 0 < e < 2} + 77.
Thus p*(r) < sup{r| — 6(e); 0 < e < 2}, because rj > 0 was arbitrary.
The dual statement is obtained similarly.
?
For a Hilbert space H we easily calculate Ph(t) — \A + t2 — 1; in
particular, H is uniformly smooth. In fact, Hilbert spaces are the "most
uniformly convex" and "most uniformly smooth" spaces; precisely, for every
Banach space X, we have 6x(s) < 1 — y 1 — ^ and px{j) > Vl + t2 — 1
(Nordlander; see, e.g., [Disl]).
Theorem 9.10 (Smulian; see, e.g., [DGZ3])
Let (X, || • ||) be a Banach space and (X*, || • ||*) its dual
(i) The norm || • || is uniformly convex if and only if || • || zs uniformly
Frechet differentiable.
(ii) The norm \\ • \\ is uniformly Frechet differentiable if and only if\\ • ||
is uniformly convex.
Proof: (i): Let || • || be uniformly convex with the modulus of convexity
5(e), and let p*(r) be the modulus of smoothness of || • ||*. Let 6q > 0
be given. By the definition of 6(e), we have 6(e) > 6(eo) > 0 for every
ee[e0,2].
Let r G @,6(e0)). For e G [eo,2], we have § - 6(e)/r < § - 6(e0)/r <
I — 1 < 0, so by Lemma 9.9,
p*(r) /e %)\ /er\ e0
——= sup (--_)< sup (d = T-
Hence lim ^- = 0.
r-0 T
Conversely, if || • || is not uniformly convex, then there is ?0 > 0 such that
6(eo) — 0. Then, from Lemma 9.9, we have for every r > 0
p\r)= sUp{f -6(e)}>?-f.
0<<-<2^ ^ J Z
Therefore limsup ?-^- > ?0, and || • ||* is not uniformly Frechet differen-
7--+0 T
tiable.
(ii): It is proved similarly using the second statement in Lemma 9.9.
?
From Theorem 9.3 and Theorem 9.10, we get the following theorem.
Theorem 9.11
Let (?l,p) be a measure space with a a-finite measure p. If p G (l,co), then
Lp(p) is uniformly smooth.

9. Uniform Convexity 291
Since duals of spaces whose canonical norms are not uniformly convex,
l\ and i1^ are not uniformly smooth. Consequently, Co, ?i, and l^ are not
uniformly smooth.
Theorem 9.12 (Milman, Pettis)
Let (X, || • ||) be a Banach space. If || • || is uniformly convex or uniformly
Frechet differentiate, then X is reflexive.
PROOF: Let (X, || • ||) be a uniformly convex space and / G Sx* be given.
Choose xn G Sx such that lim (f(xn)) = 1. Then, given e > 0, for n,ra
ra—>-oo
greater than some no we have 2 > ||xn + ^m|| > /(#n + ^m) > 2 — 6:. Hence
{#n} is Cauchy by the uniform convexity of || • || and lim [xn) — x for
n—*-oo
some x ? Sx- Clearly, f(x) = 1, so / attains its norm. By Corollary 3.56,
X is reflexive.
Assume now that the norm is uniformly Frechet differentiable. Then the
norm of X* is uniformly convex by Theorem 9.10. Therefore, X* is reflexive
by the first part of this proof and thus X is reflexive.
?
Finite Representability
Definition 9.13
Let Xj Y be Banach spaces. We say that Y is crudely finitely representable
in X if there is K > 0 such that for every finite-dimensional subspace
FofY there is a linear isomorphism T of F onto T(F) C X such that
||r||-||r-1||<A:.
We say that Y is finitely representable in X if for every e > 0, Y is
crudely finitely representable in X with constant K — 1 + e.
We observe that if Y is finitely representable in X and W is finitely
representable in Y, then W is finitely representable in X. The notion of
finite representability was first studied by James [Jam5], [Jam6].
Theorem 9.14
(?) ?2 is finitely representable in every infinite-dimensional Banach space.
(ii) Every Banach space is finitely representable in c$.
(Hi) Every Banach space is finitely representable in (^^00J-
(iv) Let X be a Banach space. Then X** is finitely representable in X.
For the proof of part (i), the Dvoretzky theorem ([Dvo]), we refer for
example to [FLM] or [T-J].
Proof: (ii) and (iii): Let Y be a Banach space, e > 0, and F be a finite-
dimensional subspace of Y. Let {?/z-}"=1 be an ?-net in Sf- Find /2- G Sy*
such that fi(yi) = 1, i = 1,. .., n. Consider the map T: F —+ i7^ defined for

292 9. Uniform Convexity
j/GFbyT(j/) = (/i(y))in=1.Then
||T(y)|| = maxd/^y)!} < maxfll/ill • ||y||} < \\y\\.
Given y E Sf, we find i E {1, •.., n} such that \\y — yz|| < e. Then
l|r(y)|| > |/i(y)| - \fi(yi) + fi(y - yo)\ > \fi(yi)\ - \\M\ • ||y - yi\\ > 1 - e.
Therefore, ||T(y)|| > A - e)||y|| for every y E F, so \\T~l\\ < j^.
To prove (ii), it is enough to consider the map ST of F into c0, where S is
a linear isometry from P^ into cq defined by S((x{)) = (x\,..., xn, 0,...).
oo
For (iii), we consider an isometry S from P^ into (]P ^0J defined
n = l
by 5((xj)) = @,.. .,0,(a?i),0,...), where (x{) is in the n-th block in
(?02-
(iv): We prove the following stronger statement due to Lindenstrauss and
Rosenthal.
Theorem 9.15 (principle of local reflexivity; [LiR2], [JRZ])
Let X be a Banach space. For every finite-dimensional subspace E of X**,
finite-dimensional subspace F of X*, and e > 0, there exists a linear
isomorphism T of E onto T(E) C X such that \\T\\ • WT^W < 1 + e,
x*(T(x**)) = x**(x*) for z* E F and x** E E, and T is the identity on
Enx.
Proof: (Stegall [Ste3]) Choose 6 > 0 so that 0F) < A + 0, where 0F) is
the function in Exercise 1.28. Choose aj,..., aj^ E Sx* containing a basis
of F and such that ||z**|| < A + E)supj |z**(ap| for all x** E E. Finally,
choose a <5-net {b\*,..., &**} in Sjs such that {b{*,..., &?*} is a basis of
?fll and {fcp,..., 6**}, r > k, is a basis of E. Let g = n — r. Then,
for every p E {1,..., q}, we have unique scalars tP}i, 1 < i < r such that
K*+P= EWr« Define
{W * < r>
-1. i = r + p,
0, r < i < n and z ^ r + p.
(m \
©I for m E N, and define A0: Xn -+ X*+? by
n n
^-o(#l> • • • > xn) — \xli • • • j ^fcj / J sl,ixii • • • > / v SqiXij .
* = 1 2 = 1
The operator Aq is onto since the matrix (sP)i) has rank g. Note that
tsj.ib** = 0 for j = l,...,g, so A*0* (b{*,..., b*n*) ? X*+*. Define

9. Uniform Convexity 293
U:Xn — Knm by U(xu...,xn) = {aj(*0}?;=i- Let Z = Xfc+* x Rnrn,
and define A: Xn -+ Z by
A(xi,..., xn) - [A0(xi,..., ar„), {7(xi,..., xn)j .
By Exercise 5.28, A is an operator with a closed range. Since A is onto and
Ao*(&i*,'..., 6^*) G Xk+q, there exist di,..., dn such that A0(di,..., dn) —
A*Q*{b\*,.. .,&**). We claim that in fact there exist ei,...,en such that
^(ci)...)en)=J4"(tr,".,0
Indeed, since U is a finite-rank operator, {/(Ker(Ao)) = J7(Ker(Ao)**) =
C/(Ker(A5*)). Because FJ*,... ,&**) G (di,. ..,d„) + Ker(A5*), there exists
(ci,..., en) e (di,..., d„)+Ker(i40) so that U(eu ..., e„) = <7(&**,..., 6**)
and the claim follows. In particular, A**F^*,..., &?*) € A(Xn). By
Exercise 3.39, there is (&i,..., bn) G X so that A(bu ..., bn) = A**F^*,..., 6;*)
and moreover sup ||6t-|| < A + 6) sup ||6**|| = 1 + 6.
Define an operator T: E -> X by T(b\*) - &f for 1 < i < r. Note that
r r
for 1 < p < q we have ]T} Spjb** — 0 and ^ sP)iA" = 0. Thus, we also have
that TF**) = 62- for r < i < n. Note that for each i we have
||TFr)|| > sup \a* (TFD) | = sup |&**(a*)| > A + tf).
3 3
Thus, ||T|| • \\T"l\\ < 1 + e by the choice of E. This completes the proof of
Theorem 9.15 and Theorem 9.14.
?
Proposition 9.16
Let the norm || • \\x of a Banach space X be uniformly convex (resp.
uniformly Frechet differentiable, resp. let X be a Hilbert space). If a Banach
space Y is crudely finitely representable in X, then Y admits an equivalent
norm that is uniformly convex (resp. uniformly Frechet differentiable, resp.
Y is isomorphic to a Hilbert space).
We will prove separable versions; the general case requires only a minor
adjustment.
Proof: Assume that || • \\x is uniformly convex, and a separable Banach
space Y is crudely finitely representable in X with constant K > 1. Let
{xn} be dense in Y. For every n G N, put Fn = spa,n{xi}?=l and let
Tn:Fn —» X be a linear operator with ||Tn|| < K and HZ^H = 1. Define a
norm || ¦ ||n on Fn by ||z||n = \\Tn(x)\\x. Because ||x||y = WT^TnWWx <
\\Tn(x)\\x and \\Tn(x)\\x < \\Tn\\ • ||z||y, we get ||x||y < ||x||n < K\\x\\Y
for every x G Fn. Extend || • ||n by 0 to Y, and by the Cantor diagonal
argument assume that a subsequence {|| • ||nfc} is convergent at each point
of {xn}. Because of the uniform equicontinuity of all || ¦ ||n on Fn's, we
have that the sequence || • || is convergent at every point of Y and its

294 9. Uniform Convexity
limit is an equivalent norm || • ||0 that satisfies ||^||y < ||#||o < i?||:c||y for
every x ?Y. We claim that || • ||0 is uniformly convex.
Indeed, from the uniform convexity of the norm || • ||^ on X, we have
that given e > 0, there is 8 > 0 such that if u,v E X satisfy ||?i|| E
A-6,1 + 6), \\v\\ E A-8,1 + 8), and ||h±* || E A-8,1 + 8), then \\u-v\\ < e.
Given x,y in Y such that ||x||o = \\y\\o = 1 and H^-^llo > 1 — §, then
for k large enough we have that \\x\\nk = \\Tnk(x)\\x E A — 8,1 + 8),
\\y\\nk = \\Tnk(y)\\x e A-8,1 + 8), \\^\\nk = ||rnfc(^)||x > 1-8, and
|lk~2/||o ~ lk-y||nfc| = |lk-y||o - ||Tnfc(x-y)||^| < §. Since from the
preceding we get ||Tnfc(z - y)\\x < ?, we have \\x - y\\0 < e.
The proof for the other cases is similar.
?
Note that it follows from Proposition 9.16 and Theorem 9.12 that if the
norm of a Banach space X is uniformly convex or uniformly Frechet differ-
entiable and Y is crudely finitely representable in X, then Y is reflexive.
Since, for instance, cq is not reflexive and is finitely representable in the
reflexive space QC^oo)o ^Y Theorem 9.14, (X^^o ^oes no^ adrciit any
norm that is uniformly convex or uniformly Frechet differentiable.
Definition 9.17
A Banach space X is said to be superreflexive if every Banach space finitely
representable in X is reflexive.
Note that X = (^^00J *s a ren<exive space which is not superreflexive.
Theorem 9.18 (Enflo [Enf], James [Jam4])
Let X be a Banach space. The following are equivalent:
(i) X is superreflexive.
(ii) X admits an equivalent uniformly convex norm.
(Hi) X admits an equivalent uniformly Frechet differentiable norm.
(iv) X admits an equivalent norm that is uniformly convex and uniformly
Frechet differentiable.
To prove this theorem, we need to introduce trees in a Banach space.
For n = 0,1,... and e > 0, an (n,?)-tree in a Banach space X is defined
inductively as follows. For n = 0, any x0 E X is a @, ?)-tree. Assume that an
(n, ?)-tree was formed by adding the points x\,..., x2^ to an (n— 1, ?)-tree,
then an (n + 1, ?)-tree results from adding 2n+1 points yi, zi, i — 1,..., 2n,
to the (n,?)-tree in such a way that X{ = \(yi + zi) and \\iji — Z{\\ > e for
each i. Therefore, an (n,er)-tree has 2° -f h 2n = 2n+1 — 1 elements.
We arrange the elements of an (n,e)-tree into a sequence as follows, xo
is the vertex. By #1,0:2 we denote the elements added at the second step,
so we have xq — (x\ + x2)/2, ||#i — x2\\ > s. By ?3, X4, x$, xQ we denote the
elements added at the third step, so we have x\ — (x^+x^)/^, ||x*3—?4|| > e,
and x2 - (x$ + x6)/2, \\x5 - x6\\ > e. In general, x{ - (x2i+i + x2i+2)/2
and ||aj2f+i — #2i+2|| > ? for every i < 2n — 2.

9. Uniform Convexity 295
An (oo,?)-tree is defined by repeating this construction countably many
times.
Note that if n E N, we can build an (n,2)-tree in B^ as follows: set
x0 = @,...,0), xx = A,0,. ..,0), x2 = (-1,0,...,0), *3 = A,1,0,...),
xA — A, — 1, 0,0,..., 0), etc. Therefore, X — (^^00J nas the property
that, for every n, Bx contains an (n,2)-tree. However, X cannot contain
a bounded (oo,?)-tree for any e > 0. Indeed, we have the following result.
Theorem 9.19
A reflexive Banach space does not contain a bounded @0, e)-tree for any
?>0.
Proof: By contradiction, assume that a reflexive space X contains a
bounded (oo,?)-tree T. Since T is separable, we may assume that X is
separable. Put C — conv(T). Then C is weakly compact and convex and
thus contains a strongly exposed point x E C (Theorem 8.28). Thus there
is / E X* such that f{x) = supc(/) = supT(/) and there is 8 > 0 such that
\\z — x\\ < I whenever z €T and f(z) > f(x) — 8. Since supT(/) = f(x),
there is t E T such that f(t) > f(x) - 8. Let i = (t1 +t2)/2, \\ti -t2\\ > ?,
^1)^2 E T. Then for one element of {^i,^}, sav ^i> we have f(t\) > I — 8.
Thus \\t — x\\ < I and ||^i — x\\ < |, and hence ||2i—?|| < |, a contradiction
21
D
with II* —ti|| = |||*i-*2|!>
Proposition 9.20
Let X be a Banach space. If there is e > 0 such that Bx contains an (n, ?)-
tree for every n E N, tAen tfAere is a Banach space Y finitely representable
in X such that By contains an {po)e)-tree.
Proof: For n E N, let {3?; i = 0,l,...,2n+1 - 2} be an (n,s)-tree in
.2n+1-2
BX- For (or,-) E c00, put ||(a;)||n = E a^
By the separability of
coo and the boundedness of {x?}, we can find a sequence {n^} in N such
that for every (a2) E coo, the limit ||(a;)|| = lim||(az)||nfc exists for every
(«f) G coo-
Denote by Y the space (coo, || • ||). Note that || • || is a seminorm. Consider
the closed subspace N = {v E Y; ||f || = 0} and put Yi = Y/N. Clearly, Yx
is a normed space that is finitely representable in X. By {e;} we denote
the images of the canonical basis of coo in Yi under the quotient map. Let
Y be the completion of Y\.
Note that if W is a Banach space and Z is a dense subspace of W, then
W is finitely representable in Z. Indeed, let e > 0 and let F be a finite-
dimensional subspace of W with a normalized basis {w\,..., wn}', let K > 0
be such that for every w — Y^aiwi we nave ^ ]C la«l < IMI < ^ E la»l-
Choose Zi E Z such that \z{ — W{\ < e/K and define an operator T: F —>
span{*,-} by T(w) = ?>,•*,• for w = ?a^-. Then ||H| - ||T(u;)||| <

296 9. Uniform Convexity
?|<*i| • |K - Zi\\ < K\\w\\ • e/K. Therefore, ||T|| • ||^T—1|| can be made
arbitrarily close to 1.
This means that Y is finitely representable in Y\, and hence in X. We
claim that {e;}z-GN forms a bounded (oo,?)-tree in Y. Indeed, given i G N,
we have ||ez-||nfc = 11-^*11 — 1> an(^ hence {e2} is bounded.
Next, given i G N, for every n such that 2n — 2 > i (that is, x™ is not
in the last row of the (n,?)-tree) we have x™ — (#22+1 + ^21+2)/^ — 0-
Therefore, for k large enough, we have ||e; — (e22-j-i + e22+2)/2||nfc = 0, so
passing to the limit with k) we get ||e2- — (e2*+i + e2«+2)/2|| — 0. This means
that e,- = (e22+i + e22+2)/2 G Yi, and hence in Y.
Similarly, we show that ||e2i+i — 622+2!| ^ e> which completes the proof.
D
Note that from Proposition 9.20 and Theorem 9.19 it follows that, given
a superreflexive space X and e > 0, there is no such that Bx contains no
(n, ?)-tree for n > uq.
For a Banach space X and xo ? X, consider a sequence {#o, • • • > a^2n+1-2}
of vectors such that for every 0 < i < 2n — 2 we have X{ = #22+1 + #22+2,
II II 2fe+1-2
Ik2i+-i|| = Ik2z+2||, and Lff^r- Jlff^flf > ^* Clearly, a:0 = ? x{ for
2=2fc—1
0 < k < n. The 2n elements {^2^-1, • • •>?2n+1-2} are then called the (n,?)-
decomposition of xq. Since ||#2i+i|| = ||^2«+2|| > 2^ ^or 0 < * < 2n — 2, we
have by induction that \\xj\\ > ||x0||/2* for every 2k - 1 < j < 2k+l - 2. It
is easy to show that {#0, 2#i, 2x2, • • • > 2nx2n+i_2}, where the coefficient of
Xi is 2k for 2k - 1 < i < 2fc+1 - 2, forms an (n,6r||a?0||)-tree in X.
If X is superreflexive and e > 0, by the note after Proposition 9.20 there
exists n G N such that for every (n, ^-decomposition {xi) of xq we have
p^-1| > 2||x0|| for some 2n - 1 < j < 2n+1 - 2. Therefore,
27l+1_2 2n+1-2 n |, i, n
^0 . /on ^N
? INI = IM+ E INI>fe=r + Bn«i)
*=2»-l i=2n-l}i^j
1^0
2n
> M +
2n
By a telescopic argument, we obtain
' 1-2
II IM>fi4- x W ii
? INI>(i + 2^r)*INI
2' = 2fc7l-l
for every (Arn, ^-decomposition of #0- Thus, if X is superreflexive and
i\, e > 0 are given, there is M — M(e,K) such that for all (M,e)-
2M 2M
decompositions z = J2 xi °^ z ? X \ {0} we have X^ I kill ^
2=1 2=1

9. Uniform Convexity 297
Lemma 9.21
Let (X, || • ||) be a super reflexive space, e G @, |). Then there is a
nonnegative function | • | on X such that:
(a) \ax\ — |a||#| for all x G X and scalars a;
W (i - f)NI < W < (i - n)NI f°r evervx e *;
(c) /Aere exists 77 > 0 smc/j tf/m/ z7 IMI — IMI — 1 a?2^ 11^ "~ 2/11 ^ ?; ^en
k + y| < 1*1 + Is/I -*?•
Proof: We define | • | for ? G X by
2n
where the infimum runs through all n G No and all (n, ^-decompositions
x — Yl uj °f x-
3 = 1
\\x\\
It is easy to check (a). To get (b), note that \x\ > ^-^ > ||ar||(l - |).
1 + 6
The @, ^-decomposition {x} yields \x\ < -^ < ||ar||(l - ^).
1+8
We now prove (c). Let M = M(e, 1 + |). By (b), we may assume in the
definition of |x| and \y\ that n < M. Fix 6 > 0 and let x = ui-f v>2-\ f-u2*,
y = vi + h v2i give approximations of \x\ resp. |i/| up to 8, kj < M.
Assume / > k. By summing up the appropriate vectors from the (/,?)-
decomposition {vi)f=l of y, we obtain a (&, ^-decomposition {wi}j=1 of y
satisfying
? IHII ? IMI
Since ||a?|| = ||?/|| = 1 and \\x — y|| > ?, we have that # -f y = ux + h
K2fc + ttfi + h w2fc is a (i + 1, ^-decomposition of x + y and thus
EIM1 + EIK1I
i + l(i-4^)
Therefore, we have
>\x + y\.
W+M-k + fl > S'M +OM
1 •" 6 V1 4*+*/ ^ 6
E1M1 + E1KII 25
i + l(i-4^)
(l + |(l-4^r))(l + |(l-4^))

298 9. Uniform Convexity
6 4^2 Z^ \\wj\\ 2X
(l+l)(l+l(l-4^))
e 1 \x\-±{\y\ + 6)
e
- 2 -"I*+2
? A ?\ ~r ,* . . 1
32-4M V6 ~ 8>
> ™(?-i)-»
* 32^ F-64)-M = ">0
for 5 sufficiently small.
?
Lemma 9.22
Let (X, || • ||) be a superreflexive space, e G @, |). Ze< | • | 6e a nonnegative
function on X with properties (a)-(c) in Lemma 9.21. Then X admits an
equivalent norm ||| • ||| which has the following properties:
(A) A - f) ||*|| < 1*1 < A - i) 11*11 for every * € X;
(B) if ||*|| = ||2,|| - 1 and \\x - y\\ > 5e then |||* + t/| < ||*|| + |t,|| - rje,
where rj is as in Lemma 9.21 (c).
Proof: For x e X, define
n
I z || = infj^l^ - Zj-i|; n G N, {a^}^ C X with x0 = 0}xn = zj.
i=i
We clearly have |||a^l = I^IHl for every ar G X and every scalar a. The
function ||| • ||| satisfies the triangle inequality: Choose any 6 > 0, and let
0 = a?o, a?i,..., a?n = a? and 0 = y0, 2/1, •. •, ym = 2/ approximate |z|| and
11/1 up to <5. Then, by considering the sequence xq, #i, ..., ?n = xn -f yo3
^n + yi, ^n + 2/2, ..., zn + 2/m for ? -f y, we get
Ilk+ 2/11 < kl -»o|+ k2-«l| + ---kn ~^n-l|
+\yi -yo\ + \y2~-yi\ + -- + \ym -ym-i\
< |x| + tyr+2?.
Since 6 was arbitrary, the triangle inequality follows. Thus, ||| • ||| is an
equivalent norm.
The sequence {0, z} gives |zj < \x\ < (l - ^)||ar||. On the other hand,
for every sequence we have
J2\*i-*i-i\ > E(i- 1)\\xj -x^w > (i - J)ii*h
and (A) follows.

9. Uniform Convexity 299
To prove (B), let \\x\\ = ||y|| = 1 and \\x — y\\ > be, and let 0 = x0, xi,
..., xn = x and 0 = ?/o> Vi, • • •> Vm — V give an approximation of ||z| and
12/1 up to 6 G @, |). Note that if x lies on the segment [x{, ?;-fi], then the
sequence x0 — x0, ..., xz- = Xi, afj+i = ?, x2-+2 = ^z+i, • • •, #n+i = #n
n n-j-1
satisfies E |a?j- — Xj-\\ = E |?j — Xj_i\. By inserting division points in
j = l j=l
this manner, we may assume \\yj; — Vj-i\\ — \\xj — Xj-\\\ for j < min(m, n).
Assume that n < m. Then we have
1 „ „ ^ x II in- - x -i .*• 1*1 + 5
1 = ||x|| < > \\x-j — a;,-_i|| < > \Xj — Xj-i\ <
— 2~i ''xi ^i — i-M — i _ i 2—/ \xi xi
A - &) + f
V 12^ 2 <1+?
1 - -
5 I-,
m
Similarly, we have 1 < J2 \\Vj ~ %"-i|| < 1 + ?•
n n m
Since ? ||*j - *j-i|| = E 112/x ~ 2/j-i||, we have ? 11%' ~ 3/j — 111 < ?•
J = l J=l J=n + 1
Hence
n n
^||(ar,-- ?;_i)- (yt- -2/*-i)|| > X)((*t - *i-i) - (y{ -t/,--i))
1=1 2=1
= |kn- 2/n|| = |k-2/n||
> \\x-y\\-\\y-yn\\>5e-e = 4e.
Denote J = {i G {l,...,n}; \\(xi - x;_i) - (yt- - t/,-_i)|| > e||xt--x,-_i||}.
Then
2X^I^" ~^'-ill ^ X^H^2' ~ x«-i)~~ B/* /»-i)ll
= X] Ufa*" Xi-J ~ (yi ~ ^¦-i)ii
2=1
- XI IK*1'" x»'-1) ~ (^' ~ 2/i)ll
2-gJ
> 4e: - ]P ||(a?i - x{_i) - (yi - t/z_i)||
n
> 4e - ]T^e||a:» ~~ ^-ill > 4e - ]T\||;r; - a?i-i||
i$J i = l
> 4e-e(l+e) >2e.
Also, if z G J, then
|(zx- - ar*_i) 4- (j/t- - s/»—1)| < |st- - s»-i| + |sfi - 2/«—11 - »7||s* - Zi-i||-

300 9. Uniform Convexity
Consider a sequence {zk} C X of N = n + m — \J\ points such that
z$ — 0, zjsf = x + y, and the differences of two consecutive points are
(a?j_i + yi-i) - (xi + yt) for i ? J, xi-1 - a?t- and yt-_i - y2- for i ^ J, and
t/j_i — 2/2- for z > n. Note that the order in which these differences occur
does not matter in the following argument. Then
Ilk+ 2/1 < ^|(^*-^«-i) + B/i-2/t-i)|+ X^ kt-art--.i|
+ ]C 12/* — S/»—11 -I- 5^ l2/» — 2/* —1|
i<n,i?J i>n
< J^ki-xi-i\ + ^2\y*~^-il~^Xllki-^-ill
+ ]l k*-^-il + XI l^-^-il + X^'"^-1!
i<n,i?J iK.n^i^J i>n
Since 77 and ? do not depend on 6> and 5 was arbitrary, we have
This proves Lemma 9.22.
?
Proof of Theorem 9.18:
(i) => (ii): For en — 2~n~3, n ? N, we find by Lemma 9.22 an equivalent
norm ||| • ||n on X satisfying (l - ^)\\x\\ < |H|n < (l - ft)\\x\\ and
7yn > 0 such that whenever \\x\\ = \\y\\ = 1 and \\x — y|| > 5en, we have
00
l*+y|» < l*|n+|y|n-i?nen. Define anorm |-|0 on X by I*|0 = E "fK
n = l
Finally, define an equivalent norm || • ||0 by ||a;||o = (||klo + Ikll2J-
We claim that || • ||0 is uniformly convex. Let xm,ym E X be such that
lim B||xm||g + 2||ym||g - \\xm + ym||g) = 0 and {xm} is bounded. Then
m—>oo
lim (||zm|o - |s/m|o) = 0 and lim (|km|| - \\ym\\) = 0. Assume that
m—>oo m—y<x>
lim ikmlo = lim |||i/mffio = 1 and lim ||zm|| = lim \\ym\\ = a > 0.
m—»-oo m—»»oo m—>oo m—»-oo
Consider the vectors um = m and um = m . It is enough to show
lkm|| Ibmll
that lim ||rzm — vm\\o = 0. Indeed, having this,
m—kx>
Ikm - awm||0 = ||km|| ~ a|||wm||o -* 0
since {t/m} is bounded. Similarly, \\ym - avm\\0 —> 0, so ||zm - ym||0 —> 0.
By contradiction, assume that \\um — vm\\o > 6 > 0 for some 6 > 0 and
all m. Fix no such that S ? [2^+3, 2n^+2 ]. By Lemma 9.22 and a convexity
argument, we have
|tim|o + |Vmlo-|t«m + Vm|o

9. Uniform Convexity 301
> 2^(lKllk + IIKIIno - ||K + «m|„0)
- 2no
This is a contradiction with the fact that |||wm|||o -* 1/a, Ill ^m III o "-*¦ l/a>
V;
and |um + vm|o = k rr + r, rr -* -. Thus, (i) implies (ii).
M
2
a
(ii) => (iii): If X admits an equivalent uniformly convex norm, then
X* admits an equivalent uniformly Frechet differentiate norm by
Theorem 9.10. By Proposition 9.16, every space Y that is finitely representable
in X* admits an equivalent uniformly Frechet differentiable norm, and thus
by Theorem 9.12, every such space Y is reflexive. Therefore, X* is super-
reflexive and by (i) =>¦ (ii) we have that X* admits an equivalent uniformly
convex norm || • ||. The predual norm to || • || on X is then uniformly Frechet
differentiable by Theorem 9.10.
(iii) => (i): Let ||| • ||| be an equivalent uniformly Frechet differentiate
norm on (X, || • ||). Let Y be finitely representable in (X, || • ||). Then Y is
crudely finitely representable in (X, |-|) and, by Proposition 9.16, Y admits
an equivalent norm that is uniformly Frechet differentiate. Therefore, Y
is reflexive by Theorem 9.12. Hence X is superreflexive.
We now prove that (i)-(iii) implies (iv). Let || ¦ || be an equivalent
uniformly convex norm on X, and let || • ||0 be an equivalent uniformly Frechet
differentiate norm on X. Define || • \\*n on X* by ||/||;2 = ||/||*2 + i(ll/IISJ>
where || • ||q is the dual norm to || • ||0. Since || • ||q is uniformly
convex by Theorem 9.10, we get that || • || are uniformly convex, and their
predual norms || • ||n on X are thus uniformly Frechet different iable by
Theorem 9.10. Define a norm on X by ||z|2 = ?2~n|klln-
Note that the derivatives of II • [L are bounded on bounded sets and thus
ii \\n
|| • || is uniformly Frechet differentiable as all || • ||n are.
If xm,ym ? X satisfy B|||zm|||2 + 2|||ym|2 - |||zm + ymf) — 0 and
{xn} is bounded, then the same is true for all the norms || • ||n, and
since || • ||n converge uniformly on bounded sets to || • ||, we have that
B||?m||2 + 2||ym||2 - ||arm + ym||2) -+ 0. Since || • || is uniformly convex,
we have lim \\xm — ym|| = 0. Hence ||| • ||| is uniformly convex.
m—>-oo
?
We refer to [Lan] for a different proof of Theorem 9.18.
Theorem 9.23 (Kadec [Kadi])
Let (X, || • ||) be a uniformly convex Banack space with modulus of convexity
CO
6(e). If^xi converges unconditionally in X, then ^ ^(ll^jll) < °°-
The proof is based on the following lemma.

302 9. Uniform Convexity
Lemma 9.24
II n
In the same notation, let z\,..., zn EX be such that max Y2 azi
ez-±l\\i=l
< 2.
Then E*(lkill)<l-
j=i
n
Proof: Assume without loss of generality that for sn — ^ Zj we have
3=1
II n ll
||sn|| > \\J2 ?jzj for every choice of Sj = ±1. Then, for jG{l,...,n} we
Hj=i I'
have by our assumption
mi<
2|MI _
ySn ZZj
s (s — 2z)
Since 8(e) is a nondecreasing function and since n and ^-^—r-^—
are both in Bx by the assumption, we have
S71 ^n ^Zj
6(\\*i\\) < S{
< 1-
Therefore,
X>(IM) < ?(
If Sn . (S„ -2Zj)>
+
= 1
i=i
i=i
!h-?^
i=i
1 II n II
< n-ii7TrE(^-^) ==
J=l
Proof of Theorem 9.23: There is n0 such that
n > no and every Sj = ±1. Therefore, ^ ^(Ikjll) < 1 f°r every n > no,
— n —
lMll(n"
,h that
1 n 1
L^i ?j xj
-l)fin|| = l.
D
< 2 for every
J=n0
and hence ]P ^(||^j||) < °°-
3=1
?
Pisier proved that every superreflexive space can be renormed so that its
modulus of convexity satisfies 8(e) > kep for some p > 2 ([Pisl]).
It is known that the moduli of convexity of Lp(p) spaces have the
following behavior: If 1 < p < 2, then there exists kp > 0 such that for the
modulus of convexity 8p(e) of Lp(p,) we have 6p(e) > kpe2, e E @,2]. If
2 < p < oo, then there is kp > 0 such that 8p(e) > kpep, ? E @,1]. We refer
the reader to [Kadi] or [Meil], [Mei2] for the proof.

9. Uniform Convexity 303
If the modulus of convexity of a Banach space X satisfies 6(e) > kep for
p > 2, then X is of cotype p\ see, for example, [DGZ3]. The canonical basis
of ?p shows that ?p is not of any cotype q < p. L\(p) is of cotype 2 (see,
e.g., [LiT3]).
As a corollary to these results and Theorem 9.23, we obtain that if 1 <
p < 2 and Y,xi *s an unconditionally convergent series in Lp([0,1]), then
Y |kill2 < oo. This is a result of Orlicz.
Recall that a Schauder basis {x^ of a Banach space X is called
seminormalized if 0 < inf \\xi\\ < sup ||ar2|| < oo.
Theorem 9.25 (Gurarii, Gurarii [GuGu], James [Jam6])
Let (X, || • ||) be a superrefiexive Banach space with a seminormalized
Schauder basis {xi}. Then there exist p, q E (l,oo) and K > 0 such that
oo
for every x = Y aixi E X we have
^(EH?)^NI<^(ENP)?-
2 = 1 2=1
Theorem 9.25 will follow from Theorem 9.18 and Lemmas 9.26 and 9.27.
Lemma 9.26
Let (X, || • ||) be a Banach space with a seminormalized Schauder basis {xi}.
If X is uniformly convex, then there is p E A, oo) and K > 0 such that for
OO / OO \ -
every x — Y aiXi in X we have \\x\\ < K [Y \ai\p)? ¦
«=i \"=i '
Proof: We may assume without loss of generality that the basis is
normalized. Fix some e E @, bcA..\) and set A — 2A — 6(e)), where 6(e) is the
modulus of convexity of || • ||. Fix p E (l,logA 2). First, we will show that
there is a E @,1) such that \\x + ty\\p < 1 -f tp whenever \t — 1| < a and
a:,yG Sx are finitely and consecutively supported vectors.
We may assume that / = max(supp(x)) < min(supp(?/)). Let Pi be the
canonical projection associated with {xi}. Then bc{^2-}||x — y\\ > \\Pi(x —
y)\\ = IWl = i-Tnus \\x - v\\ > tclxi} > e>and we nave \\x + v\\ ^ ^-Since
A? < 2, we have \\x + 1 • y\\p < 1 + lp.
Since the function ||x+/y||p is uniformly continuous at t = 1 with respect
to x, y E Sx, we have that there is a E @,1) such that \\x -f ty\\p < 1 -f tp
for \t — l\ < a and xy y as stated.
/ n \ i
Set K = I". It is enough to show that ||zn|| < K • [Y la*'lP J f°r everv
\'=i '
n
finitely supported vector zn = Y ®ixi- We proceed by induction on n E N.
i=l
For n = 1, the result clearly holds. Suppose that it is valid for n. Consider
n + l
a nonzero vector zn+\ = ^ cxiXi. We distinguish two cases:
2 = 1

304 9. Uniform Convexity
A) \ai\ < ?||zn+i|| for all i = 1, 2,..., n + 1;
B) |aio| > ^||zn+i|| for some i0 G {l,...,n + 1}.
s
In the first case, put zq = 0, zs ~ J2 &i%i for s = 1, 2,..., n + 1, ys =
2 = 1
n + 1
J2 ociXi for s - 0,1,..., n, yn+i = 0. Then ||z0|| < ||y0||, lkn+i|| >
2 = 5 + 1
ll^n+ill, |lkf+i|| ~ IH|| < ^Ikn+iH, and |||y,-+i|| - ||ft||| < jHl*M-i|| for
i — 1,.. . ,n.
Claim
Let {&}?_]_ and {^}"=1 be two sets of real numbers and € > 0. Assume that
fi < m, ?n > 7/n; and |6+i-&| < e, |r/i+i-77i| < e/or i = 1,2,... ,n-1.
Tften tfAere zs an index z'o G {1,..., n} swcA that |ft-0 — »/*0| < s.
This follows by considering the first index j for which ?j > rjj.
By this claim, there is r G {1,... ,n} such that |||^r||"~||2/r||| < iflkn+ill-
By interchanging zr and yr if needed, we can assume that ||zr|| > j|t/r||- By
the homogeneity, assume without loss of generality that ||zr|| = 1. Since
\\Vr\\ < \\zr\\ andzn+i = zr+yr) we get ||2„+i|| < 2; so |l-|M|| < j[ - <*.
Put t = \\yr\\ and yr — t^-u- Applying the first part of this proof to
x — zr and y — yr (note that |1 — t\ < a) yields \\zr + tyr\\p < 1 + tp.
Therefore,
\\Zn + l\\P = \\Zr+yr\\P = \\Zr+tyr\\P < 1 + *" = ll*r||P + Hlfrll'-
By the induction hypothesis,
ikrir+iij/rir<A^^Kr+^ fl np-
2*=1 z=r+l
n + 1
Therefore we have ||zfc+i||p < Kp ^ \oti\py which concludes the first case.
2 = 1
In the second case, assuming |a2-0| > ^||^n+i||, we get
n + 1
zn+1\\<K-\aio\<K-(jr\ai\py.
1=1
Therefore, the induction step to n + 1 is justified and Lemma 9.26 is proven.
?
Lemma 9.27
Let (X, || • ||) be a Banach space with a seminormalized Schauder basis {#;}.
// || • || 25 uniformly smooth, then there is q G (l,oo) and L > 0 such that
OO • OO \ -
for every x = J^ ociXi in X we have \\x\\ > -M ?2 \oti\q) .
i=i \=i '

9. Uniform Convexity 305
Proof: Let fi be the biorthogonal functional to X{. Then {/,-} is a
Schauder basis of X* because X is reflexive by Theorem 9.12. We have
\\fi\\ > j^?- > sup||^H» WM\ ^ 2bc{a?,-}, so {fi} is a seminormalized basis.
Since the dual norm || • ||* of X* is uniformly convex by Theorem 9.10,
we can use Lemma 9.26 for the Schauder basis {fi} of X*. Therefore, there
oo
is p G A, oo) and K > 0 such that for every f — Yl Pifi ? ^* we nave
2 = 1
ii/ir<A'.(?>f)p.
i-l
of Lemma 9.27. Indeed, let zn = E a2x2- and let gn = E fafi E X*. Then
n
We claim that L — j^ and the dual index q = -?-j- satisfy the conclusion
lei
^ffgfi1 < |bn|| < K • (? |Af ) *. Therefore,
1 I / M 1 2-v *** Pi
I, M>J_ IgnfcQI _ 1 lj=l 1
(EIAI")' (E^V
Choose /% = |or2-1^^ sign (a2) for i = 1,..., n. Then we have
n p
, EN* , n i
M^Tf—^(Ekl')'-
(SKI*-1) 2-1
\=i 7
D
Exercises
9.1 Show that a norm of a finite-dimensional space is uniformly convex if
it is strictly convex.
Hint: Use the compactness argument.
9.2 Show that a norm || • || of a Banach space X is uniformly smooth if and
only if for every e > 0 there is 6 > 0 such that ||a: + y|| + \\x — y\\ < 2 + e\\y\\
whenever ||z|| — 1, ||y|| < 8.
Hint: Directly from the definition.
9.3 Let norms || • ||n of Banach spaces Xn be uniformly convex with moduli
of convexity 6n(e), n G N. Assume that 6(e) — inf(<5n(?)) > 0 for every

306 9. Uniform Convexity
e > 0. Show that then the canonical norm of (XX Xn, || • ||n)J is uniformly
convex.
Hint: Direct calculation.
9.4 For n G N, let || • ||n be a strictly convex norm on P^ such that
II-Hoc < IHI„ < 2H-IL- Let X = (E(CIHI„)J- Show that X is
locally uniformly rotund.
Hint: Direct calculation.
9.5 Let X be a uniformly convex infinite-dimensional Banach space whose
modulus of convexity satisfies 6(e) > kep for some k,p > 0 and all ?@, 2].
Use Theorem 9.14 (iii) to show that p > 2.
9.6 Let the norm || • || of a Banach space X be uniformly convex (resp.
uniformly Frechet differentiable). Assume that Y is a closed subspace of X.
Show that the canonical norm of X/Y is uniformly convex (resp. uniformly
Frechet differentiable).
Hint: (X/Y)* is isometric to the subspace YL of X*. Use Theorem 9.10.
9.7 Let X be a uniformly convex space with modulus of convexity 6(e).
Take / G Sx* and consider the affine hyperplane H — (/_1@) + z) for
some z eX. Show that if dist@, H) > 1 - 6(e)/2, then diam(Bx nH)<e.
Hint: If x, y G Bx C\ H} then (x + y)/2 eHC\Bx.
9.8 Let X be a finite-dimensional Banach space. Show that if a Banach
space Y is finitely representable in X, then Y is isometric to a subspace of
X.
Hint: First, note that dim(Y) < dim(X) and that the unit ball of the space
of operators on an n-dimensional Banach space is compact. If Tn: Y —> X
have the property that both ||Tn|| and HT^1!! tend to 1, take their limit
point.
9.9 Show that if Y is crudely finitely representable in X and X has type
p (resp. cotype p), then Y has type p (resp. cotype p).
Hint: Direct examination of definition.
9.10 Show that ?q is not crudely finitely representable in ?p for q < p < 2
or 2 < p < q.
However, the Dvoretzky theorem (see, e.g., [T-J]) gives that ?2 is finitely
representable in every Banach space.
Hint: Use the preceding exercise and the best types/cotypes of ?p.
9.11 It is known that the space ?\ has cotype 2 (see, e.g., [LiT3]). By
looking at the standard unit vectors in Co, we see that cq does not have
cotype 2. This gives that c0 is not crudely finitely representable in ?\. Find
an elementary proof that cq is not finitely representable in ?\.

9. Uniform Convexity 307
Hint: (James) First, show an auxiliary elementary statement that if a, 6, c >
0, then the sum of four terms \a ± b ± c\ is at least |(a -f b + c) (consider
a > b > c > 0). Consider now x,y,z G Si1 and use this fact on each
coordinate of the four vectors x ± y ± z to see that the sum of the norms
of these four vectors is at least 5. Therefore, at least one of the vectors
x ±y±z has norm at least |. This cannot happen in cq (use the standard
unit vectors).
9.12 Let X be a superreflexive space. Show that if Y is isomorphic to X,
then Y is superreflexive.
Hint: If |]| • ||| is a uniformly convex norm on X (Theorem 9.18) and T is
an isomorphism of X onto Y, then the norm ||| • ||i defined for y G Y by
IyI! = |T"B/)||| is an equivalent uniformly convex norm on Y.
9.13 Show that if || • || is a uniformly smooth norm of a Banach space X,
then (|| • || y is uniformly continuous on bounded sets.
Hint: Direct calculation, the chain rule, and use the fact that the first
derivative of the norm is bounded.
9.14 Let X be a Banach space. Show that X is superreflexive if and only
if X admits an equivalent norm || • || such that the derivative (|| ¦ || )' is
uniformly continuous on a neighborhood of the origin.
Hint: One direction: see the previous exercise. If (|| • || )' is uniformly
continuous on a neighborhood of the origin, then the Minkowski functional
of the appropriate level set of || • || gives an equivalent uniformly Frechet
differentiable norm on X.
9.15 Let X be a superreflexive Banach space with a normalized Schauder
00 x °° x
basis {xi\. Show that the series V" — converges in X but Y] _ . n „x
n=l ** n^iln(n + l)
does not converge.
Hint: J^ n~p < oo for every p > 1. Use Theorem 9.25.
If the second series did converge, then Y" —-z— < oo for some q G
lny(l + n)
A, oo) by Theorem 9.25, which is not the case.
9.16 A norm || • || of a Banach space X is called uniformly rotund in
every direction (URED) if for every z G Sx and all bounded sequences
{xn},{xn} C X such that 2||zn||2 + 2\\yn\\2 - ||xn + yn\\2 -> 0 and
xn — yn — Xnz for some An, we have An —> 0.
Let T be an uncountable set. Show that co(T) has no equivalent URED
norm.
Hint: Let |] - Hoo ^e ^ne suPremum norm of co(r), and assume that || • || is
an equivalent norm on c0(r). Set M = sup ||a?||. Let xn G co(r) satisfy
lklloo<i

308 9. Uniform Convexity
||a?n||oo = 1 and ||xn|| —> M. Let z be such that ||z||oo = 1 and its support
is disjoint from all the supports of xn. Then \\xn ± z||oo = 1 for every n.
Since ||(xn + z + xn — z)/2\\ —> M, we have \\xn ± z\\ -* M and also
2||xn + z\\2 + 2||a:n - z\\2 - \\xn + z + xn - z\\2 -+ 0. Thus || • || is not URED.
9.17 Let r be uncountable. Use the preceding exercise to show that there is
no bounded linear one-to-one operator from Co(T) into any URED Banach
space X.
Hint: Let T:co(T) —> X be such an operator. Let || • || be an equivalent
URED norm on X, and define an equivalent norm ||| • ||| on cq(T) by ||x||2 =
\\x\\lo + ||T(x)||2. Let {xn}, {yn} be bounded sequences in c0(T), z G c0(r)
be such that \\zWoq = 1, xn-yn — Xnz for some An, and 2|a?nl2 + 2|yn|2 —
Ikn + 2/n|||2 -> 0. Then, as in Chapter 8, we have 2||T(xn)||2 + 2||T(yn)||2 -
ll^C^n + yn)\\2 —> 0. Since || • || is uniformly convex, we have T{xn — yn) —
XnT(z) -> 0. Since T{z) ^ 0, we have An -h- 0. Thus ||| • ||| is URED, a
contradiction with the previous exercise.
9.18 Show that every separable Banach space has an equivalent URED
norm.
Hint: X* is w*-separable; use the proof of Theorem 8.19.
9.19 Show that the space t\ does not contain any bounded (oo,?:)-tree for
any e > 0.
Hint: Norm closed bounded sets in separable duals have strongly exposed
points (see Chapter 8).
9.20 Since i\ is not superreflexive (it is not reflexive), there is e > 0
such that B^ has arbitrary large (n,?)-trees. Show this fact by a concrete
example.
Hint: Build a D, l)-tree as follows. Split (|, |, |, |,0,...) into the pair
(i0,^,i,0,...) and @,|,I,I,0,...). Then split (|, 0, |, |,0,...) into
A,0, |,0,...) and (§,0,0,1,0,...), etc.
9.21 Let X be a Banach space. Assume that every separable closed sub-
space of X admits an equivalent uniformly convex norm. Show that X has
an equivalent uniformly convex norm.
Hint: Assume the contrary. Then there is e > 0 such that for every n there
is an (n,^)-tree in Bx> The union of these trees lies in a separable closed
subspace Z of X. Then Bz contains (n,?)-trees for all n, and hence Z is
not superreflexive.
9.22 Let {Kn} be a sequence of nonempty compact convex subsets of a
topological vector space such that Kin U i^2n+i C Kn for all n. Show that
then there exists an infinite tree {xn} such that xn ? Kn for all n.

9. Uniform Convexity 309
Hint: Consider K — Y[Kn. For n G N, define
An - {X - {xk} G K\ Xn = ~(x2n + #2n+l)}-
Note that each An is closed and hence compact in the compact set K and
that x — {xn} G K is an infinite tree if x G f] An. Thus, we mustonly prove
that p| An ^ 0. By compactness, it suffices to show that AiC\A2n.. .C\Ak ^
0 for every &. To this end, fix k and define x G if as follows. If n > k, let
xn be an arbitrary element in Kn. From n = k to n = 1, we proceed by
induction: if for m = n + 1, n + 2,... points zm G A'm were chosen, define
xn — \{%2n + «2n+i)- The resulting ar is in A\ n A2 0 ... n A*.
9.23 Let X be a Banach space. Show that if X contains a separable closed
subspace Y such that Y* is nonseparable, then there exist e > 0 and a
bounded set A in X* such that every nonempty relatively u>*-open subset
of A has diameter greater than ?.
Hint: By the proof of Theorem 8.9, there is a subset C in Y* that is
unclosed and contained in By*- Let R be the restriction map of X* onto
Y*. Note that i2 is u;*-^*-continuous and maps Bx* onto .By*. Let A be
a minimal u>*-compact subset of Bx* such that R(A) = C. Let U be a
relatively w*-open subset of A. Then Ci = R(A\U) is a w*-compact set in
C, which is a proper subset of C due to the minimality of A. Then C\C±
is a relatively w*-open set in C, and thus it has diameter greater than e.
Since R is a 1-Lipschitz map, it follows that diam(i7) > e.
9.24 (Stegall [Stel]) Assume that a Banach space X has a separable closed
subspace Y of X such that y* is nonseparable. Show that X* contains a
bounded infinite ?-tree for some e > 0.
Hint: (van Dulst-Namioka) Let A be the set from the preceding
exercise. Construct a sequence {Un} of relatively w*-open subsets of A and
a sequence {xn} G Sx such that U2n U U2n+i C ZTn for every rc and
(#* — 2/*)(#n) > ? for every x* G i72n> V* G E^n+i, and n. We construct it
by induction. To see the first few steps, put U\ = A. Since diam(Ai) > e,
there are Zq and z\ in U\ such that \\z$ — z\|| > ?. Choose a point a?i G ?x
such that (zq — zl)(xi) = ? + 5 for some <5 > 0. Let
U2 = {** G f/i; s*(*i) > z5(xi) - E/2},
% = {</* G C/i; y*(*i) < zj(ari) + 5/2}.
We use the following observation: if 3\ and J2 are sets in a
topological vector space, then conv(Ji) — conv^) C conv(Ji — J2). Indeed,
a — conv(J2) — conv(a — J2) C conv(Ji — J2) for every a G Ji- Having
the sets Un constructed as above, define Kn = conv™ (Un) for every n. We
have i^2n U K2n+1 C i^n for every n. By Exercise 9.22, there is a tree {xn}
in X* such that xn G ivn for each n. By an earlier remark, we have
(X2n ~ *2n + l) € (#2n ~ #2n+l) C COnV^E^n ~ ^2n + l),

310 9. Uniform Convexity
so we have (x*2n - x*2n+1)(xn) > e. Hence \\x*2n - a?5n+1|| > e and x*n is a
bounded infinite e-tree in X*.
Note that it follows from this result that if X is a separable space with
nonseparable dual, then there is a bounded convex norm closed set C in X*
that contains no strongly exposed point. Indeed, the norm closed convex
hull of the tree in StegalPs result is such a set.
9.25 (Daugavet) Show that if A is a compact operator in C[0,1], then we
have ||/ + A\\ — 1 + ||^4|| (we say that A satisfies the Daugavet equation).
Hint: Assume that A is a one-dimensional norm-one operator x «—> f(x)y,
where y G C[0,1] and / G C[0,1]* corresponds via the Riesz representation
to a function <p with finite variation. Let xo G Sc[o,i] satisfy ||A(a?o)|| > 1—?•
Let 2/0 = A(x0). Let J be an interval where y0 is greater than 1 — ? and
the variation of (p on J is small enough. Change xo to x\ to have xq = #i
outside J, #i(^o) = 1 for some to ? J: and a?i is a broken line on J. Put
t/i = -A(si). Due to the smallness of the variation of cp on J, we have
||yo -Vi\\< e- Then ||/ + A\\ > Xl(t0) + yi(tQ) > 1 + 1 - 2e.
9.26 Use the Daugavet property to show that C[0,1] does not have an
unconditional basis.
Hint: (Kadec) Assume C[0,1] has an unconditional basis. Let a —
sup||P/i||, where Pa are the canonical projections onto {ez}2-Gi4 for finite
A C N. Pick A such that ||Pa|| > & — \- Then, by the previous result,
||7-^|| = 11/11 + 11^11 = 1+ ||^|| >l + a-| = a + i
On the other hand, ||7 — Pa\\ — ||-Pn\a0|| ^ a- Hence a > a + |, a
contradiction.
9.27 ([AAB]) Let X be a uniformly convex Banach space and T G B{X).
Show that T satisfies the Daugavet equation if and only if ||T|| lies in the
approximate point spectrum of T.
We recall that A is a point of the approximate point spectrum of T if
there is a sequence {xn} C Sx such that ||T(xn) — Aa?n|| —+ 0.
Hint: If ||T|| is in the approximate point spectrum of T, choose {xn} C Sx
such that ||||T||xn -T(xn)\\ ~> 0. Then
\\i + t\\ > ||(z-i-r)(*w)|| > ||^ + lirii^H - lllirii*^ - r(or„)||
= 1 + ||T||-||||T|K-T(xn)||.
Now let T satisfy the Daugavet equation. Then 5 = T/||T|| satisfies the
Daugavet equation and ||/-|-5|| = 1-|-||5,|| = 2.
Hence there is a sequence {xn} C Sx such that \\xn -\- S(xn)\\ — 2. Since
\\xn\\ = 1 and ||S'(xn)||||xn|| < 1, we have from the uniform convexity of X
that ||5(x„)-a:n|| -> 0. This reads \\T(xn) - \\T\\xn\\ -+ 0, so ||T|| is in the
approximate point spectrum of T.

9. Uniform Convexity 311
In the following exercises, we will construct and investigate the Tsirelson
space T ([Tsi]). All notation and definitions made in one exercise apply to
the following exercises as well.
Let {ef} be the canonical basis of Co; we will use x(i) for the z-th
coefficient of x G Co; that is, x — ^x(z)e2-. For n G N, define a tail projection on
/CO \ CO
Co by Pn ( ]P x{i)ei J = ^ x(i)ei. Let {v{} be a finite sequence of vectors. If
\"=1 ' i-n
they are consecutively supported (which will be denoted by vi < ... < t>n),
n
we write (vi,..., vn) for ?^ v2-. When we use (vi,..., vn), it is automatically
i=l ^
assumed that Vi are successively supported.
Let A be a subset of Co. We consider the following set of conditions:
A) A C BCo and {e,-} C A;
B) if x — Y^, x{i)ti ? A and |y(z)| < |e(«)| f°r a'l z> then ]C 2/B)e« ? ^5
C) if vi < ... < v„ G A, then \Pn{(vx,..., vn)) e A;
D) for every x ? A there is n G N such that 2Pn(x) G A
9.28 Show that there is a weakly compact set K C Co satisfying (l)-D).
Hint: Put A\ - {ac,-; \a\ < 1, i G N} and
An+i =i4nu{|PjWr((t;i>...,t;iV)); ^ G N, vi < ... < v^ € A„}.
CO
Let K be the pointwise closure of (J An. Only D) needs an expla-
n = l
nation. Assume that r; G AT \ .Ai, u(z'o) ^ 0, and t?(z) = 0 for i < z'o-
CO
Choose a sequence {v1} in (J An pointwise convergent to v. We may
n = l
assume that vl(j) — 0 for all i and j < z'o and that vl (? A\. Thus
t;' = \PNi {(v\, • • -j^jV;))) an(i nence ^' < 2o- By passing to a
subsequence we may assume that v% = |Pjv((^i, • • •, vzn)) for some N G N.
Thus v — |P/v((^i5... ,vjv)), where every ty is a pointwise limit of
{vj}z- Put M = max{z; t>2- ^ 0}, ra — min(supp(t>M)) • Then by B),
Pjv((vi ,..., vN)) G /i, so 2PN(^) G A'.
The weak compactness of K follows because K is pointwise
sequentially compact. This can be seen by estimating the distribution of those
coordinates of x G K that are larger than a given e > 0.
9.29 Show that V = conv(K) also satisfies A)~D).
Hint: Only C) and D) need an explanation. To see C), consider X{ —
ol\x\ -f ... + <*"#", where x\ G K, for m > i the supports of all x\ precede
the supports of all xlm1 and x — (xi,... ,xn). Then ^P^(x) is a convex
combination of \Pn{J^i^3\ , • • •»»in4n)) > so \^n{x) ? V.
To see D), set Dn = {# G K\ 4Pn(z) G A}. Clearly, Dn C Dn+i and
A' = (J Dn. For every xq G V, there exists a Borel measure // in the

312 9. Uniform Convexity
weak topology of V such that f(x0) — f f dfi for / E Cq. Thus, for n0
K
large enough, we have fi(Dno) > |. We claim that 2Pno(xQ) E 7. Indeed,
otherwise, consider the set W = {# E V; 2Pno(z) E V}. Choose / E Cq
such that /(x0) > 1 and \f(x)\ < 1 whenever # E W. Then
Kf(x0)= J fdp+ J fdp<^(Dno) + 2p(K\Dno)<± + ± = l,
Dno K\Dno
a contradiction.
9.30 Let T = span(V) (span taken in c0) and consider the norm given on X
by the Minkowski functional of V. Show that T is a reflexive Banach space
with an unconditional basis {ez}. Show that c0 is finitely representable in
every infinite-dimensional closed subspace of T. In particular, T contains
no isomorphic copy of Co, ?i, or a super reflexive space.
Hint: T is a Banach space since V is weakly compact in cq (Theorem 3.58).
By B) and D), it follows that {e,-} is an unconditional basis of T. The basis
{e,} is shrinking. Indeed, suppose that ||P^(/)|| > 2e for some / E T*,
e > 0, and n E N. We choose x\ < x^ < • • • such that f(x{) > e. Since P/v
does not affect #jv+l + ... 4- X2N and thus H^at+i + ... + ^2Ar|| < 2, we get
a contradiction with the fact that /(zjv+i + ... + x2n) > Ne.
Having shown that {e,-} is shrinking, we show that V is weakly compact
in T. Indeed, span{e*} is dense in T*, so the ti;-topology of T restricted to
V coincides with the ^-topology of Co restricted to V, which is compact by
Krein's theorem.
If {xi} is a normalized block basic sequence of {ez}, then |P/v(#i + ... +
xN) E V for every N by C), Thus, by B), \\PN(X1x1 + ... + XNxN)\\ <
2 max | Xi \. Consequently,
max |Aj| < ||Aia?iv-M + • • • +A;v^2Ar|| < 2max|A2-|.
Using the technique from Theorem 6.20 we find that c0 is crudely finitely
representable in every infinite-dimensional closed subspace of T. Thus no
infinite-dimensional closed subspace of T can be superreflexive, and hence
it does not contain subspaces isomorphic to ?p for p E (l,oo). Since T is
reflexive, it does not contain an isomorphic copy of cq or t\.

10
Smoothness and Structure
Definition 10.1
Let X be a Banach space, n E N. A function M:Xn —> R is called an
n-linear form on X if it satisfies
M(xi,..., xjb_i, ax + /?y, xjfc+i,..., xn)
= aM(xu ..., a?fc-i, x, xk+1, ...,xn) + f}M(xu ..., ar^i, y, Sfc+i, • • •, *n)
for every k E {1,. .., n}; x, y, X{ E X and a, /? E R.
If M((Bx)n) is bounded, then M is called bounded.
The form is called symmetric if M(#i,..., xn) = M(x^i),..., x^n^) for
every permutation ir of {1,..., n}.
It is routine to check that M is bounded if and only if it is continuous
on Xn, and the linear space C(nX) of all bounded n-linear forms on X,
endowed with the norm ||M|| = sup{|M(zi,..., xn)\\ X{ E Bx], is a Banach
space.
By ?5(nX) we denote the closed subspace of C(nX) that consists of all
symmetric forms. Note that Cs(lX) = C(lX) = X*.
Multilinear forms allow us to define polynomials on Banach spaces.
Definition 10.2
We say that P:X —> R is a polynomial of degree n E N if for i — l,...,n
there are bounded i-linear forms B{ such that for all h E X we have
P(h) = fii(A) + B2(h, A) + • • • + Bn(h,..., h).

314 10. Smoothness and Structure
Basic facts about multilinear forms and polynomials can be found, for
example, in [Bar].
We define higher order derivatives by induction, using the Frechet
derivative in case n — 1.
Definition 10.3
Let U be an open set in a Banach space X, n > I, and assume that f:U—>
R is n-times differentiable in U. Suppose that the n-th derivative f(n': U —>
Cs(nX) is continuous at Xo ? U. If there exists M E ?5(n+1X) such that
hmf&H*o+th0){hl h )-/(")(xo)(ftl hn) _ M{h K)) = 0
and the limit is uniform for ho,..., hn ? Bx, we say that M = /^n+1^(^o)
is the (n + l)-th derivative of f at Xo.
We say that f is Fn-smooth (resp. Cn-smooth) on U if f^ exists (resp.
f(n^ is continuous) at all points of U.
We say that f is C^-smooth on U if it is Fn-smooth on U for all n E N.
Note that if a function / is Fn+1-smooth on some open set [/, then it is
Cn-smooth on U.
Variational Principles
Definition 10.4
A function (p: X —> R is called a bump (function) on X if it has a bounded
nonempty support.
Fact 10.5
Let X be a Banach space. If X has a Cn-smooth norm, then X admits a
Cn-smooth bump.
If X has a C°°-smooth norm, then X admits a C°°-smooth bump.
Proof: Consider any C^-smooth real-valued function r on R such that
r([— |, |]) = 1 and r(t) ~ 0 for r > 1. The composition function <p(x) =
t(\\x\\) is then a Cn-smooth (resp. C°°-smooth) function on X such that
(p@) ~ 1 and <p(x) = 0 whenever x ? Bx (see, e.g., [Die]).
?
In particular, if a Banach space X has separable dual, then there exists
a Lipschitz C1-smooth bump on X. Indeed, by Theorem 8.19, X admits an
equivalent Frechet-smooth norm || • ||, which is then Cfl-smooth onl\{0}
(Corollary 8.5).
Similarly, using the result that every separable Banach space has an
equivalent Gateaux differentiable norm, we get that every separable Banach
space admits a Lipschitz, Gateaux differentiable bump function.

10. Smoothness and Structure 315
Haydon constructed a nonseparable Banach space X that has a C°°-
smooth Lipschitz bump function but does not admit an equivalent Gateaux
differentiable norm ([Hay2]).
Theorem 10.6
Let X be a Banach space. If'<p is a continuous Gateaux differentiable bump
function on X7 then spanj^x); x ? X} — X*.
Proof: Define ip \ X -+ RU {+00} by ij){x) = <p~2(x) if <p(x) / 0 and
ip(x) = +00 otherwise. Given / ? X* and e > 0, the function ip — / is lower
semicontinuous and bounded below on X) and thus, by Theorem 3.53, there
is xo ? X such that, for every h ? X and t > 0,
il>(x0 + th) - f(x0 + th) > iJj(xo) - /(a*) - et\\h\\.
Hence, for h ? X and t > 0,
j,(z0+th)-*(*o) > /(«o+tt)-/(«o) _ ?||A|| = f{h) _ ?m
Because ip(xo) ^ 00, we have <p(xq) ^ 0, and ip is thus Gateaux
differentiable at xq. Hence, from the latter inequality it follows that, for all
hex,
i\) (x0)(h) = hm_ - > f(h) - e\\h\\.
Thus ^\x0){h) - f(h) > -e\\h\\ for all h ? X. Considering ±h, it follows
that \i/>'(x0)(h) - /(A)I < e\\h\\ for all h e X, and hence ||^(^o) - /|| =
H-R^F^o)-/||<^
D
Azagra and Deville proved ([AzDe]) that if X* is separable, then X
admits a C1-smooth bump 6 such that {b'(x)\ x ? X} = X* (c/.,
Corollary 3.56).
Theorem 10.7 ([FWZ])
Let X be a Banach space. If X admits a UF-smooth bump function, then
X is superreflexive.
Proof: Let B% denote the open unit ball of X. Let r be a Lipschitz
C^-smooth function on the real line R such that r@) = 0, r(l) = —1,
r(R) = [0,-1], and let <j> be a symmetric UF-smooth bump function on
X such that <f>' is uniformly continuous on X, 0@) = 1 and <j>(x) = 0
whenever ||e|| > |. Set 0 = r o 0. Then 0 < 0, inf@) = — 1 = 0@), and
supp@) C \B%. Note that 2||ar|| - 2 < 0(z) for all x ? 5*. For * > 0,
define
r0(? + /i) + 0(;r-/i) - 20(z) _,,,,. 1
W(t) = supj^ ; .. ; ^; x e x, ||A|| < t).

316 10. Smoothness and Structure
Define ip: B$ -+R by
n
ij){x) — miV^OLi^Xi)] ^2<XiXi = x.oci > 0,^a2- - l,a?t- G B,n G N j.
2 = 1
Note that i/j is convex. Let G — {x G ?§; ^(ar) < — |}. We claim that ip'
is uniformly continuous on G.
To see this, let x eG,h? X with 0 < \\h\\ < \} and 0 < e < -\-ij)(x) be
given. Let xi G B^, a2- > 0 with ]P a2- — 1 be such that J^ a2</>(a:2) < -0(x) +
?||A|| < — \, and assume that ^(z;) < 0 for i = 1,..., k and <j>(xk+i) = ... =
fc A: A;
0(*„) = 0. Then -\ > ? a,-tf(*,-) > - ? a,-. Thus 1 > a =?>,•> |
2 = 1 2 = 1 2 = 1
and \\x{ + ~/i|| < ||a?i|| + 2||A|| < 1 for i = 1,..., k. Consequently,
ip(x + h) + i/>(x -h)- 2ip(x)
< J^otrfixi + ±h) + ]T ai<j>(xi - ±A) - 2 J2 <*i<Kxi) + 2e||A||
2 = 1 2 = 1 2 = 1
= ?<*,•(*(*,¦ + ^/i) + 0(x,- - ift) - 2<j>{xi)) + 2e\\h\\
2 = 1
<w(JIM)llM + &||fc||.
Since O 0 was arbitrary, V>(* + ft) + V»(x - h) - 2i/j(x) < 2a;B||A||)-||/i||.
This shows the existence of ipf(x). We will now show that ipf is uniformly
continuous on G. To this end, let x} y G G, 0 < \\x — y\\ < | be given. Then,
for h G X with ||A|| = ||s-y||, we have x + he B%, y - (x + h - y) G B%,
and, by convexity,
ty'(*W(i/)HO
< rP(x + h) - V-(x) - i{>'(y)(h)
- 4>(x + h)- ip(y) - il>'(y){x + h-y)+ ^(y) - tp{x) + i/>'(y)(x - y)
< rP(x + h)- rp(y) - ^'{y){x + h-y)
< i>(y + {x + h- y)) + il>(y-(x + h-y))~ 2xl>(y)
< 2wB||* + h- y||j||x + h - y\\ < 4wD||x - y\\)\\h\\.
By taking the supremum over all h G X, \\h\\ — \\x — y\\, we obtain that
\\il>'(x) - *p'(y)\\ < 4u>D||z - y\\) for x, y G G and 0 < \\x\\ < |.
Let Q = {x G 5^; V>(x) < — |}, and let q be the Minkowski functional of
Q. If q(x) = 1, then %l)'(x){x) > ^(a?) — ip@) — \. By the implicit function
theorem, it follows that the derivative of the norm |ar| = q(x) -f q(—x)
is uniformly continuous on the sphere. The statement now follows from
Theorem 9.18.
D

10. Smoothness and Structure 317
Recall that if / is a function on a Banach space X with values in R U
{+00}, then by the domain of / we mean Dom(/) = {x ? X] f(x) < 00}.
The epigraph of / is defined by Epi(/) = {(x,t/)GXxR; y > f(x)}.
We say that / is proper if there is x ? X such that f(x) < 00; that is,
Dom(/) ^ 0.
We now state the Deville-Godefroy-Zizler version of the Borwein-Preiss
smooth variational principle ([DGZ2], [DGZ3]; c/., [BoPr]).
Theorem 10.8 (Smooth variational principle; see, e.g., [DGZ3])
Let X be a Banach space that admits a Frechet differentiable (resp. Gateaux
differentiable) Lipschitz bump function. Then, for every lower semicontin-
uous and bounded-below proper function f on X and for every e > 0,
there exists a function g that is Lipschitz and Frechet differentiable (resp.
Gateaux differentiable) on X and such that sup{\\g(x)\\] x ? X} < e,
suplH^7^)!!; x ? X} < s, and (/ — g) attains its minimum on X.
Proof: We will consider the case of Frechet differentiability; the proof for
the Gateaux differentiability case is similar. Let Y be the normed space
of all Lipschitz and Frechet differentiable functions h on X normed by
IhI = ||A||oo + ||A'||00¦ It is standard to show that Y is a Banach space. For
a given lower semicontinuous and bounded-below function / defined on X,
consider the set
Un = <g EY] there is Xq ? X such that
(/ - 9)(x0) < inf {(/ - g)(x); x?X\B° (x0, ?)}},
where 5^(a?o, ~) is the open ball with radius ^ centered at xq.
We claim that Un is an open dense subset of Y. Indeed, Un is open in Y
as ill' III > II ¦ lloo- Fix any ^ ? y and e > 0. We must find h ? Y, \\h\\ < e,
and xq ? X such that
(f-g-h)(x0)<M{(f-g-h)(x); x € X \ B°{x0, i)}.
By using the assumption on the existence of the Lipschitz, Frechet
differentiable bump function on X and the shift and multiplication arguments,
find b ? Y such that 6@) > 0, ||6|| < e, and b(x) = 0 whenever ||z|| > ?.
Since / — g is bounded below, we can find xq ? X such that
(/ - g)(x0) < inf{(/ - g)(x); x € X} + 6@).
Let h(x) = b(x - x0)- Then h € Y and |/i||| < e. Moreover,
(f-g- h)(x0) = (/ - g)(x0) - 6@) < inf(/ - g).
If x e X\B°(x0, i), then (f-g-h)(x) = (f-g)(x) > Mx(f-g). Hence
g + h ? Un, and this shows that Un is dense in Y. Consequently, by the

318 10. Smoothness and Structure
Baire category theorem, G — f] Un is a dense G$ subset of the Banach
n>l
space Y.
We claim that if g G G, then (/ — #) attains its minimum on X. Indeed,
for n > 1, using the definition of Un, find ?n G X such that
(/ - g)(xn) < inf {(/ - g)(x); xeX\B° (xn, ±) }.
We have ?p G B^[xn, ~) for p > n; otherwise, by the choice of xn) we
would have (/ - g)(xp) > (f - g)(xn). But then ||arn - xp\\ > ? > ± and
considering the choice of xp, we would then have (/ — g)(xn) > (f — g)(xp),
a contradiction. Hence, {xn} is Cauchy in X and thus converging to some
Xqo G X. We will show that (/ — g) attains its minimum on X at a^.
By the lower semicontinuity of (/ — #), we have
(/-flO(*oo) < liminf(/-#)(xn)
n—>-oo
< Hminf(inf{(/-^)(x); x e X\B$(xni±)})
< inf{(/-^)(x); xGl\W}.
The last inequality follows since if y jL x^, then \\xn — y\\ > ^ for large
n, and so for such n inf{(/ - flf)(z); x E X\B^(j:n^)} < (/ - g)(y).
Therefore, (/ — g) attains its innmum on X at .x'oo.
D
Corollary 10.9
If a Banach space X admits a Lipschitz, Frechet differentiable (resp.
Gateaux differentiable) bump function, then for every convex continuous function
f on X, the set of all points of Frechet (resp. Gateaux) differentiability of
f is dense in X.
In particular, if a Banach space X admits a Lipschitz, Frechet differentiable
bump function, then X is an Asplund space.
There is a Banach space X and a convex functions f on X such that the
set of all points of Gateaux differentiability of / is dense but not residual in
X (see, e.g., [Fab]). On every nonseparable reflexive space, there are convex
continuous functions whose set of points of Gateaux differentiability is not
Gs ([HSZ]), although it must be residual (Theorem 8.26).
It is not known whether every Asplund space admits a Lipschitz, Frechet
differentiable bump function. In particular, it is not known whether C(K)
admits a Frechet-smooth bump if K is the Kunen compact (see Remark
after Corollary 12.52).
It is also an open problem whether a Banach space X is weak Asplund
if every continuous convex function on X is Gateaux differentiable on a
dense set in X. (In this context, it would be interesting to check the so-
called two-arrow space; see Chapter 12.) However, it is known that X is a

10. Smoothness and Structure 319
weak Asplund space if it admits a Lipschitz, Gateaux differentiate bump
function (see [DGZ1]).
Proof of Corollary 10.9: We will present the proof for the case of
Frechet differentiability. The proof for Gateaux differentiability is similar.
Let /be a concave continuous function on X. Let 6 be a Lipschitz and
Frechet differentiate bump function on X with 6@) / 0 and b(x) = 0
for x ? Bx- Let xq ? X. Choose 8 > 0 such that f{x) > f(xo) — 1 for
x ? Bx(xo, 8). Choose m > ~. If \\x — xo\\ > 8, then \\m{x — xo)\\ > 1 and
b(m{x — xq)) = 0. Define a function cp on X by
( \ — f b{m{x — xq))~2 if b(m(x — x0)) ^ 0,
^ ' ~ \ -fco otherwise.
Consider f + (p. Then f + (p = +oo outside B^{x^y 8) and is bounded below
by f(xo) — 1 on Bx(xq, 8). Hence / + <p is bounded below on X.
It is easy to check that f + <p is lower semicontinuous. By Theorem 10.8,
there is a Lipschitz, Frechet differentiable function g on X such that / +
(p — g attains its minimum at some point yo (that necessarily belongs to
B^(xoi6)). On some neighborhood U of yo, the function b(m{x — xq)) is
not zero, so <p is Frechet differentiable on U. For y ? ?/, we have /(t/) +
^(y) ~ 9(y) > /(yo) + ?>(yo) ~ ^B/o) and thus
0 < (-/)(yo + A) + (-/)(yo-A)-2(-/)(yb)
< ^(y0 + A) + <Kyo - A) - 20(yo)
-y(yo + A) - y(yo - A) + 2</(y0) < o(||A||) + 0(||A||)
for small h. Thus, / is Frechet differentiable at yo (see Chapter 8).
?
The following open question is related to variational principles. If / is a
convex continuous function on Co, it is not known whether there are Xq ? cq
and K, 8 > 0 such that f(x0 + A) + f(x0 - h) - 2f(x0) < K\\h\\2 for all
he6BCo.
Theorem 10.10 ([HaSu], [Ste4])
Let X be a Banach space. If X admits a Lipschitz, Gateaux differentiable
bump function, then (Bx*,™*) is sequentially compact.
Proof: Let {/n} be a bounded sequence in X*. For n ? N, put An —
w
{fj}j>n and A = f] An. Define a function on X by p(x) — sup {/(#); / ?
A}. By Corollary 10.9, p is Gateaux differentiable at some point x0 ? X.
By Smulian's lemma, we obtain that there is a unique /o ? A such that
f0(xo) — p(xq). Because f0 is in all An, for each j there is fnj ? {/,}
such that \fnj(xo) — fo(xo)\ < j- Let fi be a w*-cluster point of {fnj}-
Then fi(x0) — f(x0) — p(x0). Moreover, fi ? A. From the uniqueness of

320 10. Smoothness and Structure
the element of A such that fo(xo) = p(#o), we have that /o = /i- By a
standard argument, fnj —¦» /o-
D
Definition 10.11
Let X be a Banach space. Let f:X —* (—oo,-foo] be proper and xq G
Dom(/). The Frechet sub differential D~ f(xo) of f at xq is defined by
r>-*( \ / r- v* r • t f(xo + h)-f(x0)-p(h) \
D /(xo) = {peX; hmmf — > 0 j.
The Frechet superdifferential D+ f(xo) of f at xq is defined by
D+f(X0) = {PeX*; limsup^+^-^-^^Q}.
1 ||*||-o \\n\\ >
We say that f is Frechet sub differentiate (resp. f is Frechet superdiffer-
entiable) at x0 if D" f(x0) / 0 (resp. D+f(x0) ^ 0).
For example, for the function f(x) = \x\ on R and x0 = 0, we have
D-f(xo) = {kt; \k\ < 1} and D+f(x0) = 0.
Note that every continuous convex function / on a Banach space X is
Frechet sub differentiate at all points of X. Indeed, by application of the
separation theorem for the sets Int(Epi(/)) and (xq, f(xo)) in I ® R, we
obtain pGl* such that f(xo + h) — f(x0) > p(h) for all /i G X. Hence
peD-f(xo).
We will use the following two facts.
Fact 10.12
Let X be a Banach space. Let f:X—± (—oo, +oo] be proper and xo G
Dom(/). If f is Frechet subdifferentiable and Frechet superdifferentiable at
xo, then there is p G X* such that D~ f(xo) = D+ f(xo) — {p} and f is
Frechet differentiable at xq with f'(xo) = p. On the other hand, if f is
Frechet differentiable at xq, then D~ f(xo) = D+f(xo) = {f'(xo)}.
PROOF: Indeed, if p G D~ f(xo) and q G D+f(xo), then for every h G Sx
we get
P(h) < liminf/(^o+^)-/(^)
^v } - *-+o t
< hmsup — —- < q(h).
t—o t
Thus (g — p) > 0 on Sx and it is a linear functional; hence q — p = 0. Thus
D~ f(xo) =: D+f(xo) = {p}] from the definition of the superdifferential
and sub differential, we get that p is the Frechet derivative of / at xq.

10. Smoothness and Structure 321
On the other hand, if / is Frechet differentiable at #o, it is Frechet
sub differentiable and Frechet superdifferentiable at xo, f'(xo) G D~ f(xo):
and f'(xo) G D+ f(xo). From the first part of this proof, it follows that
D-f(x0) = D+f(xo) = {f'(xo)}.
?
Fact 10.13
Let f,g be functions on a Banach space X. Ifg is Frechet sub differenti able
at xq G X, f(xo) = g(xo) < oo, and f(x) > g(x) on a neighborhood of xo,
then f is Frechet sub differenti able at x$.
Proof: Let p G D~g(x0). Then
liminf f{xo+h)-f(xo)-p(h) > lim.nf g(xo + h)-g(x0)-p(h) > Q
Hfcll-o ||A|| " ||*||-o \\h\\
D
Theorem 10.14
Let X be a Banach space X that admits a Lipschitz, Frechet differentiable
bump. If f is a proper lower semicontinuous function on X, then the set
of all points of Frechet sub differenti ability of f is dense in Dom(/).
The proof is similar to that of Corollary 10.9, and we omit it.
Note that we cannot in general expect that the set of points of subdiffer-
entiability contains a dense G$ set in Dom(/), even for Lipschitz functions.
Indeed, otherwise we would obtain (by intersecting sets of sub
differentiability and superdifferentiability) the Frechet differentiability on a dense G$
set, which is not true (see Chapter 8).
Definition 10.15
Let X be a Banach space, and let f:X —> (—oo,+oo] be a proper function.
For x G Dom(/), we define the (Moreau-Rockafellar) sub differential df(x)
of f as
df(x) = {z* G X*; f(x + h) - f(x) > x*(h) for all h G x\.
If / is convex and continuous, then from Lemma 8.4 it follows that /
is Gateaux differentiable at x if and only if df(x) is a singleton; then
df(x) consists of the Gateaux derivative f'{x) of / at x. By the separation
theorem, we get that df(x) / 0 if / is convex and continuous at x.
Let / be a proper function on a Banach space X. The (Legendre-Fenchel)
conjugate f*:X* —> R to / is denned by
/*(**) = supK(z) - /(*); x G X).

322 10. Smoothness and Structure
Thus /* is a convex function from X* to (—00, +00]. We also put /** =
(/*)*. Ifg:X* -> (-00,+00], we define #*:X-+R by
g*(x) - s\ip{x*(x) - g(x*)\ x* E X*}.
Lemma 10.16
Let f:X—> (—00,4-00] be a proper function on a Banach space X. Let
x E Dom(/) and x* E X*.
(i) x* E df(x) if and only if f*(x*) + f(x) = x*(x).
(a) Ifx* E df(x), then x E df*(x*) and f**(x) = f(x).
(Hi) If f is convex on X} then f**\x = f if and only if f is lower
semicontinuous.
(iv) ([BeMo]) // / is convex and lower semicontinuous, then Epi(/**) =
Epi(/) and df(x) — df**(x)f)X* forx E X. Moreover, if f is continuous
at x, then /** is also continuous at x.
Proof: (i) and (ii): Let z* E df(x). Then
/*(**) = sup{z*(y)-/(y); yeX}
= -/(*) + x*(x) - inf{/(y) - f(x) - x*(y - *); y E X)
= -f(x) + x*(x)m
If this identity holds and y E X, then
f(y)-f(x) = /(y)+ /*(**)-**(*)
> f{y) + x*(y)-f(y)-x*(x) = x*(y-x),
so x* E df(x). Then, for y* ? X* also
w)-r 00 = ran+/(*)-**(*)
> »» - /(*) + /(z) - x'(x) = (y* - *•)(*),
so x E df*(x*). The identity also yields
/oo = *»-r(an <r*(*).
The reverse inequality follows from a general fact that /** < /. Indeed, for
z ? X we have
r*W = suP{y*(z)-r(y*); 2/* ex*}
< sup{2/*B) - (y*(z) -/(z)); y* G X*} = /(z).
(iii): Assume that / is convex and lower semicontinuous. Then the
separation theorem guarantees that df(x) / 0 for every x E X where f(x) < 00,
and thus/**(*) = /(x) by (ii).
(iv): First, assume / > 0. Then Epi(ff C Epi(/**) follows from
Epi(/) C Epi(/**) and the u;*-lower semicontinuity of /**. Let (xq* , A0) E
Epi(/**). Suppose that (xq*,Ao) ^ Epi(/) . By the separation theorem,

10. Smoothness and Structure 323
there is Xq E X* and k) a, C E R such that:
aS*(*o) + k\0<a<C< x**(xl) + k\ for all (x**,\) E Epi(/) .
From these inequalities we get k > 0 (if k < 0, it is enough to take x E
Dom(/) and A —» oo in order to obtain a contradiction). In particular, we
get x*0) + kf(x) > /? for all x E Dom(/). Take e > 0. Since / > 0,
_^oW _ w x < _^_ for all x G Domm
fc + e k + e
hence /•(-,?)< -&. Then
/••«•) > 4-(-^-)-r(-rf-)>^-(-j|-) +
fc + e/ V k + eJ ~ v Ar -f- ey k + e
1:0»-*rw))>/,"a+4Ao
If fe = 0, then /**(^o*) > (P ~ <*)/e. Since ? > 0 was arbitrary, we get
^o* ? Dom(/**), a contradiction. If k ^ 0, since ? > 0 was arbitrary, we
get f *(z**) > (p-a + k\0)/k > A0. This contradicts D*,A0) E Epi(/**).
Now, if f:X —> RU {+00} is an arbitrary semicontinuous convex
function, choose x$ E Dom(/*). Consider g:X —> RU {+00} given by
^r(x) = /(xj) + /*(#o) — ^o(x)- This function is proper, lower
semicontinuous and convex. Moreover, Dom(/) = Dom(</) and # > 0. Observe that
g**(x**) = f**(x**) + /*(x*) - x**(x*0) for all z** E X**. By the first part
of the proof, the proposition holds for </, and hence for /.
df{x) — df**(x) Hi* is a consequence of (i) and (ii).
To prove the last assertion, assume that / is continuous at xq and /**
is not. Let Wbea basis of uAopen neighborhoods of 0 in X**. Then there
exists e > 0 such that, for every U E U, we can choose x*? E xo + U
such that \f**(xtf) - f(x0)\ > e. Because Epi(/)" = Epi(/**) and /**
is lower semicontinuous, it is possible to choose xjj E (xq + U) D X such
that /**(xj>*) < f(xu) < /**(*{/) + e/2. It follows that xv ^ xQ and
\f(xu) — f(xo)\ > e/2, a contradiction.
?
Given a Banach space (Z, || • ||), a proper function f:Z —+ (—00,-1-00],
z0 E Dom(/), Zq E Z*, and Z > 0, put
<*(/,*<>, *?,*) = sup{/(z)-/(zo)-z*(z-z0); z E Z, ||z - z0|| = t},
PifyZOyZO't) = 'mHf(Z)-f(Z0)-4(Z~Z0)] Z ? Z ,\\z - Z0\\ = t} .
Lemma 10.17
Let f:Z —> (—00,-f-oo] be a proper function on a Banach space Z. If zq E
Dom(/) and Zq E df(z0), then /?(/*, zj, z0, s) > 0 an J
aC/^Oj^O + W^o^o,*) > te /or a//*, s > 0.

324 10. Smoothness and Structure
/// is moreover Frechet differentiate at zq, then /?(/*, Zq , zq, s) —> 0
implies s —> 0.
Proof: /?(/*, Zo,z0,s) > 0 follows immediately from zj E df(z0).
Take s > 0 and consider z* E Z* with ||z* — Zq|| = s. Using Lemma 10.16
for z* E Z*, we can estimate
/*(**) -/*K)-(^-^o)(^o)
= sup{z*(z) - /(z); z E ^} + /(*<,) - 2*(*o)
= -inf {(/(*) - f(zQ) - z*0(z - z0)) + (z0* - z*)(z - z0); z E Z}
>-M{a(f,zo,zl\\z-z0\\) + (z*-z*)(z-z0)] z E Z}
= -inf{a(/,z0)z*,*) - ||z0* - z*\\t; t > 0}
= sup{-Qf(/,2r0,2r5,t) + ^; * > 0},
and the inequality follows.
Assume that / is Frechet differentiate at zq and there are A > 0 and
s,- > A, i E N, such that /?(/,^o,^o>5«) < I for aU z ? N- Tnen 7 >
/?(/, zo, Zq, Sj) > sz-tf — a(/, zo5 ^o j0 f°r a^ ^ > 0- Hence, if sq is a cluster
point of the sequence (s;), we have so > A > 0 and sot — a(/, zo, Zq , t) < 0
for alH > 0. This is impossible, because the Frechet differentiability of /
at zo implies that lim(a(/, zo, z^t)/t) = 0.
?
Lemma 10.18
Let f:Z —» (—oo, -f-oo] be a proper lower semicontinuous function on a
Banach space Z such that inf(/) > —oo. // the conjugate function f* is
Frechet differentiable at Zq E Z*, then the derivative z$ — (/*)'(zo) belongs
to Z and /**(z0) = /(z0).
Proof: We note that for t E R\ {0} the functions
are w*-lower semicontinuous and for t —> 0 converge to zo uniformly on
Bz+- Thus, zo is tu*-lower semicontinuous there. Since zo is linear, zo is
w*-continuous by Theorem 4.44 and z0 E Z.
In order to prove that /**(zo) = f(zo), for z E Z put
^(z) = inf j J2 OLif{zi)\ z = 22 otiZi.OLi > 0, z{ E Z, i = 1,..., m,
i=1 t=i
m
^fti = l,mE Nj.

10. Smoothness and Structure 325
Note that tp is a convex function minorizing /. Because /** | is the supre-
mum of all affine continuous functions minorizing /, we have /** L <
First, we show that (zo, /**(zo)) is in the closure of Epi(^). Assume that
this is not true. Since Epi(^) is a convex set, by the separation theorem
there are (?, s) G Z* x R and c, d G R such that
f (*o) + sf*(z0) <c<d< ?(z) + st for all (z,t) G Epi(^). (*)
Take z G Dom(/). Then (z,t) G Epi(^) for all large t G R, so (*) implies
that s > 0. Assume for the moment that 5 = 0. Then (*) yields ?(z — zq) >
d — c (> 0) for all z G Dom(/). For n G N, define an (affine continuous)
function on Z by
0n(*) = -n?(z - zq) + n(d -c) + inf(/).
Observe that gn minorizes / and hence also /** | . Thus, in particular,
f**(zQ) > lim (gn(zoj) = lim (n(d - c)) + inf(/) = +oo.
n—*oo n—»-oo X
However, by Lemma 10.16, f**(z0) = *3(zo) — f*(zo) G R, a contradiction.
Therefore s > 0. From (*) we then have
/(*) > K(^o - *)+/**(*<>) + Krf-c)for a11 * ^ s.
The function on the right-hand side of this inequality is affine and
continuous. Thus, f(z) can be replaced by f**(z), and we get that f**(zo) >
f**(*o) + j(d ~ c), which is impossible. We have proved that (z0, f**(zo))
belongs to the closure of Epi(^).
We are ready to prove that f(zo) < f**(zo). Fix A > 0. From the lower
semicontinuity of / at zq} we find 0 < 6 < A such that f(z) > f(zo) — A
whenever z G Z, \\z — z0\\ < 6. By Lemma 10.17, there is 0 < 7 < A such
that /?(/**, zq, Zq,s) < 7 implies s < 6. Since (z0, /**(zo)) is in the closure
of Epi(^), there is (z,tf) G Epi(^) such that
t- f**(zo)- 4(z~ *o) <7-
m
Find m G N, a2- > 0 and Z{ ? Z, I < i < m, such that ]T a2- = 1,
m m / m x
? afZf = z, and ? <*i/(z») - f**(z0) - *5 ( ? a^' ~ zo) < 7- Tnen tnere
i=l 2 = 1 Vi=l '
is i G {1,..., m} such that /(*,-) - /**(z0) ~ 2^B:; - zo) < 7-
Hence /?(/**, z0, z^ II*. - *o||) < 7, so ||z,--zo|| < 6. Thus
/(*(>) < /(^) + A<7 + /**(zo) + z0*(zf-z0)-f A
< 2A + r*(z0) + ||*ol|A.
Letting A -> 0+, we get /(z0) < /**(*o).
D

326 10. Smoothness and Structure
Lemma 10.19
Let Z be a Banach space. Let g: Z* —» (—oo,+oo] be a proper lower
semi continuous function such that inf(^) > -co. If g*:Z —> (—oo,+oo]
is Frechet different!able at z0 G Z and we denote Zq — (g*)'(zo), then
gD) = (9*TD)-
Proof: Similar to the proof of Lemma 10.18 with the roles of Z and Z*
interchanged. Instead ofZxR with the norm topology, consider Z* x R
with the u>*-topology.
?
Theorem 10.20 (Stegall [Ste2])
Let Z be an Asplund space. Let <p: Z* —> (-co, +co] be a proper lower
semicontinuous function such that inf(<?>) > — oo. If liminf y4-rr > 0, then
z* II^H-^oo \\z II
for every e > 0 there are zq G eBz and Zq G Z* such that zo G cty?(zj).
Moreover, cp(z*) -<p(zq) -z0(z* — Zq) > j(\\z* — Zq\\) for all z* G Z*, where
j: @, -foo) —» [0, +oo] is such that s —>• 0 if j(s) —» 0.
In particular, <p(z% + h*) + <p(z% - h*) - 2cp(z%) > 0 for all h* G Z*.
PROOF: For z G Zt put f(z) = sup{z(z*) - <p(z*); z* G Z*}. Then f:Z-^
(—oo, +oo] is a convex function, which is bounded on a neighborhood of the
origin. Indeed, from the assumptions there are a, 6 > 0 such that <p(z*) >
a\\z*\\ whenever z* G Z* and \\z*\\ > 6. Then f(z) < max@, 6||z||-in%)).
The function <p is proper, so / is locally bounded below. / is also convex
and continuous on a neighborhood of 0. Since Z is an Asplund space, /
is Frechet differentiate at some z§ G Z with ||z0|| < min(a,?). Put Zq —
f'(zo). Then Zq G df(zo), and in the notation of Lemma 10.17 we have
/*(**)-/*(*o )-*<>(** -zS) > ftr,zS,zo,\\*'-4\\) ^ all z* G Z*. Put
-y(s) = /?(/*, ^5,zo, 5), s > 0. Then, by Lemma 10.17, 7E) —» 0 implies
s -» 0. Moreover, <p > f* and that ^(^J) = /*(^o) by Lemma 10.19. Hence
?>(**) - ?>(*S) - *o(** - *o*) > 7A1** ~ *S||) for all z* G Z*.
D
We will now show a result related to variational principles.
Proposition 10.21 ([WhZi])
Let X be a Banach space that does not contain any isomorphic copy of cq.
For every bounded closed convex set C in X, there is a compact set K C C
such that there is no h G X \ {0} with K ± h C C.
This property of K should be compared with the notion of an extreme
point.
Proof: Assume that such K C C does not exist. Choose x\ G C. Then
there is x'2 G X \ {0} such that ±xi ± x2 G C. Among all such x2, choose
x2 such that ||a?2|| > ^ sup{||^2l|; ±x\ ± x2 G C}. Having chosen x2,
consider the set {±x\ ± x2}. There is a:3 G X \ {0} such that ||a?3|| >

10. Smoothness and Structure 327
|sup{||#3||; ±x\ ±X2±x'z G C} and ±x\ ±x2 ±x3 G C. By induction, we
obtain a sequence {xi} in X\{0} such that S = < ^2 e%xu ei ~ ±1> n G N >
is in C and hence is bounded. Since X contains no isomorphic copy of Co,
J2xi is unconditionally convergent by Theorem 6.38 and the set S is
relatively compact in X. By our assumption, there is h G X \ {0} such that
S±h C C. Since this h could have been used in all steps in constructing Xi,
we have \\xi\\ > |||/i|| > 0 for all i, contradicting the convergence of ^X{.
?
Theorem 10.22 (see, e.g., [DGZ3])
Let X be a Banach space that does not contain any isomorphic copy of cq.
Let U be a bounded symmetric open subset of X containing the origin. Let
f be a continuous function on U such that /@) < 0, f(x) = f(—x) for
all x G U, and m — inf{/(#); x G dU} > 0. Then there are a compact
symmetric set K C U and A > 0 such that K + ABx C U and for every
6 G @, A) there is a finite subset K§ of K such that
inf {/*,(*); x G 7,6 < ||x|| < A} > /*6@),
where fK6(x) — sup{/(# -f &); k G K&}.
Proof: Note that since K& is finite, fx6 is a continuous function.
Put xo = 0 and, if #o, • • • , xn were chosen, set Kn — < J2 ?ixi\ ?i — ±1 f
and
En = {xeX; x + k G U and f(x + k) < f (l - ?) for all ib G A'n}.
Note that i?n is a symmetric set. Set an — sup{||^||; x G En} and choose
xn+i G En such that ||a?n+i|| > ^f-
Once the sequence {#n} is constructed, we define if = (J ifn. Then
n>0
K CU, and thus if is bounded. Because X does not contain any isomorphic
copy of Co, it follows from Theorem 6.38 that K is compact and xn —¦»¦ 0.
By the choice of {xn}, also an —> 0. By the continuity of / on [/, we get
/(>) < f for aU ^A; and hence dU n K = 0. Therefore, X C C/ and
from the compactness of K we get A such that K + ABx C J7.
Now fix any S G @, A). Choose n G N such that c*n < 6. If x is such
that 6 < \\x\\ < A, then by the definition of an and using an < <5, we have
/K-n(a?) > f A - ^r). On the other hand,
fKn@) = sup(/) = max{/K-n_1(arn),/K-n.1(-arn)} = fKn^(xn)
< f(l-5^r).
By combining these inequalities, we get fKn{x) > fKn(fy + 2^rr f°r all x
with 6 < 11a?11 < A, and the statement follows.
?

328 10. Smoothness and Structure
Theorem 10.23 ([FWZ])
Let X be a Banach space that admits a bump function f with a locally
uniformly continuous derivative. If X does not contain any isomorphic copy
of Co, then X is superreflexive.
Proof: (Sketch) Applying the previous theorem to /, U = Bx, and 8 > 0,
we obtain Kg, Consider the function
*(*)= ?('(* + *)--to)J ¦
yeK6
Then ^@) = 0 and ^(x) > e2 on some {x G X] 6 < \\x\\ < /?}. Let r be a
C^-smooth function on the real line such that r@) — 1 and r(t) = 0 for
t > %-. Then put
9{*) = \
and use Theorem 10.7.
if llxH < 6,
if ||z|| > 6,
?
Smooth Approximation
We will now pass to smooth approximation in separable infinite-dimensional
Banach spaces. One of the tools used here is the method of smooth
partitions of unity. We restrict ourselves to the case of Frechet differentiability
of the first order. Other cases of smoothness can be treated similarly.
A locally finite partition of unity {ipa} is a collection of real-valued
functions i\)a > 0 on a Banach space X such that for every x E X there is a
neighborhood U of x so that all but a finite number of i/ja vanish on U, and
^ipa(x) — 1 for every x E X. We say that a (locally finite) partition of
unity {t/>a} is subordinated to an open cover U of X if for every a there is
Ua EU such that supp('0a) C Ua, where supp(^a) = {a;El; ^a(#) ^ 0}.
We say that a Banach space X admits Frechet-smooth partitions of unity
if for every open cover U of X there is a locally finite partition of unity
{ipa} subordinated to U and such that all the functions ipa are Frechet
differentiate on X. Considering ^'a — z/>^ /^^L we can assume 0 <
ipa < 1 for all a.
Theorem 10.24 (see, e.g., [DGZ3])
Let X be a Banach space. If X* is separable, then X admits a Cl-smooth
partition of unity.
Proof: Since X* is separable, X admits a C1-smooth bump function (see
note after Fact 10.5). Therefore, by considering a shift, a squeezing oper-

10. Smoothness and Structure 329
ation, and taking squares of functions, we see that, given x G X and an
open set V C X with x ? V, there is a C1-smooth function / on X such
that / > 0 on I, f(x) > 0, and supp(/) C V. Therefore, every open set
V in X is the union of sets of the form {x G X\ fp(x) > 0}, where fp are
nonnegative Cfl-smooth functions on X. Hence, if {Ua} is an open cover
of X, there is an open cover {Zp} that consists of the sets of the form
{x G X] f{x) > 0}, where / are Cfl-smooth functions on X, such that
{Zp} refines {Ua}] that is, each Zp is contained in some Ua.
By the Lindelof property of X, there is a countable sub cover {W{} of
{Zp}. For every j, Wj = {x G X] fj(x) > 0} for some nonnegative C1-
smooth function fj on X. Put V\ = {x G X\ fi(x) > 0} and, for n > 2,
set
• Vn = {x?X; /„(*)> 0,/i(*) <?,...,/„_!(*) <?}.
For every x G X, there is n(#) such that fn(x)(%) > 0 and fi(x) —
hi?) = ••¦ = fn(x)-i(x) = 0 if n(x) > 2. Then x G Fn(»)- Therefore,
{14} is a cover of X which refines {Wj}. Moreover, {Vn} is a locally finite
cover of X. Indeed, given a?El, choose again n(x) as before. Then, choose
P € @,/n(ao(s)) and Put ^x = {y ? X] fn(x)(y) > P} • Then Qx is a
neighborhood of x and if m > max{i, n(#)}, then Vm C\ Qx — 0. To see
this, observe that by the definition of Vm, fn(x)(z) < —¦ < /? for every
z EVm and thus Fm H fi^ = 0 for such rn.
Choose now a C1-smooth function <p\ on R such that <pi (t) > 0 for t > 0
and <?>i = 0 otherwise. For n > 2, choose a Cfl-smooth function on Rn such
that y?n(*i,.. -,*n) > Ofor^j < ?, i < n, and tn > 0, and y>n(*i,- •-,<n) = 0
otherwise. For every r, put #r = <?v(/i, • • • > /r)- It follows that gr is a C1-
smooth function on X and {# G X; gr{x) > 0} = K. for every r. Put
<7 = Yl9r and <?>r = gr/g for r G N. Given # G X, all but a finite number
of (pj vanish on a neighborhood Qx and g(x) > gn(x) > 0. Thus g is a C1-
smooth function, and {(pj} is a locally finite C1-smooth partition of unity
that is subordinated to the cover {Ua}-
D
Corollary 10.25
Let X, Y be Banach spaces such that X* is separable. Lei Q, be an open set
in X. If f:Q —> Y is a continuous map, then for every e > 0 there is a
Cl-smooth map g:Q —>Y such that \\f(x) — g(x)\\ < e for all x ?&.
Proof: We will prove it for the case of Q, — X. Let V be a cover of Y formed
by all open balls in Y of radius §. Then U = f~l{V) = {f-\V)] V G V}
is an open cover of X. Let {ipa} be a Cfl-smooth partition of unity on X
that is subordinated to U. We may assume that no ipa is identically zero
on X. For each a, choose xa G X such that ipa(xct) i=- 0. Since {if>a} is
subordinated to U, if rl>a(x) / 0 for some a and some x G X, then xa

330 10. Smoothness and Structure
and x belong to the same U G U and therefore ||/(#a) — /(#)|| < e by the
definition of U.
Define a function g: X —*• Y for x ? Q by g{x) — ^ }'(xa)ijja(x). Since
{?/><*} is a locally finite partition of unity and all ij)a are C1-smooth, g is
C1-smooth on X. For x ? X, we have
\\g(X) - f(x)\\ = l52f(x°)M*)-f(*)\
= 1 E f{xa)^oc{x) ~ f(x) E iW
- I E (/(^)-/w)^w|
{a; V«(*)*0}
< E ii/(*«)-/(*)iiiM*)
{a; V«O)^0}
< e E ^aM^E^W =^
D
For more information in this direction, see, for example, [DGZ3] or [BeLi].
Corollary 10.26
Let X be a Banach space such that X* is separable. If A, B are closed
disjoint subsets of X, then there is a C1-smooth function <p on X such that
ip — 0 on A and (p — 1 on B.
Proof: Let /(*) = dist(x%^t2(x ,B) ¦ By Corollary 10.25, there is a C1-
smooth function g on X such that \f(x) — g(x)\ < \ for all x E X. Let r
be a Cfl-smooth function on R such that r(x) — 0 whenever |a?| < \ and
t(x) = 1 whenever | < x < |. Then (p — r o g satisfies the requirements.
?
Remarks
(i) The first result on a smooth approximation of continuous functions on
infinite-dimensional Banach spaces was obtained by Kurzweil ([Kur]). He
showed that if a real separable Banach space X admits a separating
polynomial, then every continuous (nonlinear) operator from X to an arbitrary
Banach space Y can be uniformly approximated by real analytic operators,
(ii) Every equivalent norm on ?2 or cq can be approximated uniformly on
bounded sets by C°°-smooth norms ([DFH1], [DFH2]).
(iii) Every Lipschitz real-valued function on a superreflexive space
can be approximated uniformly on bounded sets by uniformly Frechet
differentiable functions (see [Cep]).
(iv) There is an equivalent norm on ?2 with a Lipschitz derivative that is not
a uniform limit on Bx of functions whose second derivatives are uniformly
continuous on X (Nemirovskii-Semenov [NeSe], VanderwerfT [Van2]).

10. Smoothness and Structure 331
(v) There is a Lipschitz mapping from Bt2 into Bi2 that cannot be
uniformly approximated by uniformly Frechet differentiate maps (Tsarkov;
see, e.g., [BeLi]).
Lipschitz Homeomorphisms
Definition 10.27
We say that a metric space (JP, p) is Lipschitz equivalent to a metric
space (Q,cr) if there is a one-to-one map <p of P onto Q such that <p and
cp~x are both Lipschitz; that is, there is K > 0 such that K~1p(x,y) <
(j((p(x)^Lp(y)) < Kp(x,y). Such a map <p is then called a Lipschitz
homeomorphism of P onto Q.
Theorem 10.28 (Aharoni [Aha])
Every separable Banach space is Lipschitz equivalent to a subset of cq.
Using Corollary 5.18, we immediately have the following result.
Corollary 10.29
Every separable metric space is Lipschitz equivalent to a subset of
Coin the proof of Theorem 10.28, we will need the following assertion.
Denote by N0 the set {0,1,2,...}.
Lemma 10.30
Jjet Xo be a subset of a separable metric space X. There exists a sequence
{Mi}?lQ of subsets of Xq such that:
A) for every x G Xq} there is i G No such that dist(z, M2) < 1;
B) for every x G X, card{i G N0; dist(x, M2) < 3} < co;
E) diam(Mz-) < 8 for every i G N0.
Proof: Let Z be a maximal 1-separated family in Xo, and let Y D Z be
a maximal 1-separated family in X with respect to inclusion. Enumerate
Y — {%};6N05 assuming without loss of generality that Y is infinite. We
put Mo = Bx{yoA) H Z, where B^(j/q,4) is the open ball centered at yo
with radius 4, and for i > 1 we put
2-1
Mi = (B$(yi,4)nZ)\\jMj.
3=0
Since {Mi} covers Z, A) follows, as any maximal 1-separated family F
in a metric space W has the property that dist(u>, F) < 1 for every w G W.
The part C) follows from the fact that M{ C 5^(^,4) for every i.
To show B), let x G X be given. There is j0 > 1 such that \\x — yj0\\ < 1.
If i is such that dist(z, M») < 3, then dist(yj0, Af,-) < 4. If i > j0 and y G Mi
is such that dist(?/j0, y) < 4, then y G IJ Mj: a contradiction. Therefore,
i=o

332 10. Smoothness and Structure
i < jo whenever dist(yj0, Mi) < 4. This gives that
cardji E N0; dist(z, Mz) < 3} < jQ + 1.
?
Proof of Theorem 10.28: (Assouad) Let X be a separable Banach
space. Fix e E X and for n E Z, consider the subset Xn of X defined by
Xn = X\B%(e,G(l)n). For n E N, consider on X the metric pn = (|)np,
where /? is the canonical metric of X. For n E Z, construct the sequence
{Mt-jn}i6No applying Lemma 10.30 to the subset Xn of the metric space
(X,pn)-
For n E Z, i E N0, and ^Gl, put /2>(z) = C(|)n - p(z,M2>)) + ,
where u+ = max{u,0}. Let e2jn denote the standard unit vectors in
^oo(No x Z) and define for x E X formally
f(x) ~ Yl fiAx)ei,n-
(*,nNNoxZ
We claim that f(x) E c0(No x Z) for x E X and
?^<H/W-/(y)He„(N.xZ)<p(*,y)
for x,yGl. Having the claim proven, it suffices to note that co(No x Z)
is isometric to Co(N).
To prove the claim, given n0 E Z, consider i, m E N0 x Z such that
fi,m(x) > 3(|) °. Then, from the definition of /«,m(^), we have ra < no,
so /i,m(^) > 3(|)n° for only finitely many integers m.
Now, for any such m, whenever i E N satisfies fiiTn{x) > 3(|) °, we get
in particular fi}m(x) > 0, and thus p(a?, M2>m) < 3(|) from the definition
of fi,m(x)' By the definition of the distance pm we have pm(x, M«jm) < 3,
and by the definition of Mjjm, there are only finitely many such i. This
proves that
card{(z» E No x Z; ?,„(*) > 3(|)n°} < oo,
showing that /(a:) E co(No x Z) for every x ? X.
Let x,y ? X: x =? y. There is n E Z such that
3-4(|r<^y)<3-4(|).(|r = 18.(l)n.
Thus, at least one of x^y is farther from e than §4(|) = 6 • (|) .
We may assume that p(x,e) > 6 • (§)* , so x E An- By the definition of
Mi)U) there is z E N such that p(x,Mz>n) < (|) . Therefore fi)Ti(x) >
3 ' (I)" ~ (I)" = 2 • (I)"- 0n the other hand, if/i.nGO > 0, then from the
definition of /;>n we have p(z, Mf jTl) < 3(|) . Because the diameter of M2>n
in the canonical metric of X is at most 8 • (|) and p(z, M2-)n) < (|) ,

10. Smoothness and Structure 333
assuming that fi)U{z) > 0, we obtain an estimate for the diameter of Mt->n
in p:
P(z, z) < p(x, Mitn) + diam(Mz>) + p(z, MiyTl)
< (§)n+8-(§r+3.(§)n=i2.(ir.
Since p(x, y) > 12 • (|) , necessarily fi>n(y) = 0. Consequently,
ll/OO - /(?/)llc„(N0xz) > MA*) - kn{v)\ > 2A)" > ^-
On the other hand, from the definition of /t->n, we have that all /2>n are
1-Lipschitz. We conclude that \\f(x) - /(y)||co(N0xZ) < p(#, 2/)-
D
Theorem 10.31 (Heinrich, Mankiewicz [HeMa])
Let X, Y be separable Banach spaces. IfX is Lipschitz equivalent to a subset
ofY*, then X is crudely finitely representable in Y*.
It is not known whether X is isomorphic to Y if X and Y are
separable reflexive Banach spaces and X is Lipschitz equivalent to 7. By
Proposition 9.16 and Theorem 10.31, we obtain the following corollary.
Corollary 10.32
Let Y be a superreflexive space (resp. a Hilbert space). If X is a
separable Banach space that is Lipschitz equivalent to a subset of Y, then X is
superreflexive (resp. a Hilbert space).
In the proof of Theorem 10.31, we will use several results.
Let X, Y be separable Banach spaces and /: X —> Y*. We say that / is
w* -differentiable at xq ? X if for every h G X the limit
w._ lim /(«o + Aft)-/(«o) = {D*f){xo){h)
X—>-0 A
exists and h i-» (D* f)(xo)(h) is a bounded linear operator from X into Y*.
The operator (D*f)(xo) is called the w*- differential of/ at xq.
Henceforth, by the Lebesgue measure on an n-dimensional Banach space
X we will mean any measure on X that is an image of the Lebesgue measure
on Rn via an isomorphism between X and Rn. Since all such Lebesgue
measures are equivalent up to a multiplicative constant, the notion of a
null set (a set of Lebesgue measure zero) is well defined. We will need the
following version of Rademacher's theorem.
Theorem 10.33 (Heinrich, Mankiewicz)
Let X be a finite-dimensional Banach space, and let Y be a separable
Banach space. Assume that f:X —> Y* is a K-Lipschitz map; that is}
\\f(x) - f(y)\\Y < K\\x - y\\x for all x,yEX. Then:
(i) The map f is w*-differentiable at almost all points x G X; if x G X is
such a point, then \\(D* f)(x)\\ < K.

334 10. Smoothness and Structure
(ii) If also \\f(x) - f(y)\\y > k\\x - y\\x for all x,y G X (i.e., if f is a
Lipschitz embedding), then for almost all x G X, (D* f)(x) exists and is
an isomorphic embedding of X into Y* with \\(D* f)(x)(h)\\ > k\\h\\ for all
hex.
Proof: We will skip a standard verification that sets and functions involved
in this proof are measurable.
(i): Let {zn} be dense in Sy- Define <pn(x) — (f(x))(zn) for every
n G N and x G X. Since (pn are Lipschitz real-valued functions on a
finite-dimensional space X, Rademacher's theorem asserts that each (pn
is Frechet differentiable almost everywhere on X (see, e.g., [Fed]). Let
(D(pn)(x) denote the derivative of cpn at x. Consider the set
W — {x e X\ (D(fn)(x) exists for every n G N}.
Then X \ W is a null set in X. Choose x ?W. Then, for every h G X and
ne N,
(DVn){x)(h) = lim f(X + XV-f{x\zn).
Since / is Lipschitz, the differential quotients are bounded, and therefore
fix + Xh) — f(x)
we find that — is ,w;*-Cauchy when A —> 0. By the Alaoglu
f(x _^_ \h\ _ f(x\
theorem, if;*-lim exists in Y*. Since (Dcpn)(x) is linear
a-^o A
i ,- . * ,- f(x + \h)-f(x) . ,. , .
for every n, the limit w -hm is linear. It is a bounded
. A~+0. A
linear operator (in h) from X into Y* because / is Lipschitz and the
quotients are thus bounded. Hence, the u>*-derivative (D* f)(x) exists for every
xeW.
We also have
\\(D*f)(x)(h)\\Y. = sup \((D*f)(x)(h))(Zn)\
sup
i^ol A )M
<K
(ii): Without loss of generality, assume that k = 1. First, we claim that,
for every fixed h G Sx, the complement of the set
Mh = {xe W; \\(D'f)(x)(h)\\Y- > 1}
has Lebesgue measure zero in X. Indeed, assume the contrary. Then there
is h G Sx and m G N such that the set
N={x€W; \\(D*f)(x)(h)\\<l-±}
has a positive measure. By the Fubini theorem, there is a line L in X of the
form L — {xo + th; t G R} for some xq G X such that the one-dimensional
Lebesgue measure of L fl N is positive. Let # be a density point of the set

10. Smoothness and Structure 335
LD N on the line L (see, e.g., [WhZy]). Then there is e > 0 such that the
Lebesgue measure of the set
A = {t ? @,e); x + the LHN}
is greater than ^(l — jKm)- Set ^ = @>6:)\^- Since \\f(x + eh) —f(x)\\y >
e\\h\\x = ?, by our assumption on / there is y ? Sy such that f(x+eh)(y) —
f(x)(y) > e(l - ±). Put <p(t) = (/(* + th))(y) for t 6 [0,e].
<?> is a Lipschitz function with the Lipschitz constant at most K. By the
Lebesgue differentiation theorem, (ff(t) exists almost everywhere on [0,6:],
I^'COI — K at a^ Pomts where it exists, and <p(e) — (p@) = J* ip'(t)dt. On
the other hand, we have <p'(t) = (D*f)(x + th)(h)(y) for t ? A. Thus,
*(! " 55?) < *>(*) ~ ?>(°) = / ^@ * = / *>'(*) rf' + / P'W *
J@,e) J A JB
< f(D*f)(x+th)(h)(y)dt+ [ Kdt
J A JB
< [ \\(D*f)(x + th)(h)\\y.dt+^
J A
a contradiction. This completes the proof of the claim.
Let O be a countable dense subset of Sx- For each h ? 0, construct
the set Mh as before. Take M = f| M&. Then X \ M has the Lebesgue
measure zero in X.
If x ? M and A ? 0, then ||(J9*/)(z)(*0||y* > 1 = ||/*||x. Since (D* f)(x)
is a bounded linear operator from X into Y*, we get ||(JD*/)(x)(/i)||y* >
\\h\\x for every h ? X.
D
Proof of Theorem 10.31: Let / be a map from X into Y* such that for
some K > 1, ?||ar - y||x < \\f(x) - f(y)\\y < K\\x - y\\x for a:, y ? X. Let
Z be a finite-dimensional subspace of X. By Theorem 10.33 (ii) applied to
/ — /| we obtain x ? Z such that (D* f)(x) is a bounded linear operator
of norm at most K with \\(D* f)(x)(h)\\ > ^\\h\\. This shows that X is
crudely finitely represent able in Y*.
?
Using special measures on infinite-dimensional separable Banach spaces
(Aronszajn, Christensen, Mankiewicz; see, e.g., [BeLi]), the statement in
Theorem 10.31 can be much improved to read that, under the same
assumption, X is isomorphic to a subspace of Y* (Heinrich, Mankiewicz
[HeMa]).
Remarks
(i) If X is Lipschitz homeomorphic to Co, then X is linearly isomorphic to
cq (Godefroy, Kalton, Lancien).

336 10. Smoothness and Structure
(ii) If X is uniformly homeomorphic to lp, p ? A, oo), then X is isomorphic
to ?p (Johnson, Lindenstrauss, and Schechtman).
(iii) The superreflexivity of the space is carried over by a uniform homeo-
morphism. This is not the case with the reflexivity or Asplund property of
the space ([AhL2]).
(iv) The Asplund property is carried over by Lipschitz homeomorphisms.
At present, it is not known whether Y is isomorphic to X if X is separable
and reflexive and Y is Lipschitz homeomorphic to X.
The reference for all these results is [BeLi] or [Ben].
Homeomorphisms
It is well known that two finite-dimensional Banach spaces are
homeomorphic if and only if they have the same dimension ([Spa]). Also,
since only finite-dimensional Banach spaces have compact balls, it
follows that an infinite-dimensional Banach space cannot be homeomorphic
to a finite-dimensional one. A Banach space cannot be homeomorphic to a
non-complete normed space (Exercise 1.62).
If a locally convex topological vector space E is uniformly homeomorphic
to a normed space, then E is normable (Exercise 4.8).
If X is an infinite-dimensional Banach space, then both Bx and Sx are
homeomorphic to X ([BeP2]). On the other hand, if X is finite-dimensional,
then Sx is not homeomorphic to Bx (Exercise 10.21).
For p ? [1, oo), Stv is uniformly homeomorphic to Si2 via the map (p(x) =
|a:12 sign (x) ([BeLi]), whereas ?p is not uniformly homeomorphic to a subset
of?2 ifp>2 ([BeLi]).
We saw in Exercise 5.58 that on the unit sphere of ?\ the w*- and the
norm topologies coincide. For p ? A, oo), an analogous result for the weak
topology follows from the fact that the ?p space is uniformly convex, so it
has the Kadec-Klee property.
Proposition 10.34 (Mazur)
If p ? A, oo), then ?p is homeomorphic to ?\.
Proof: Let p ? (l,oo). Consider the map <pm.Six —> Si defined for x —
(xi) ? Stx by (p{x) — (sigYi(xi)\xi\r). Then (p is a bijection of Six onto Sip
such that <p and <?-1 are coordinate-wise continuous. On S^ and Sip, the
coordinate wise topology and the norm topology coincide, and hence (p is a
homeomorphism of Stx onto Stp.
We define a map <p:?i —+ ?p by (p(Q) = 0 and <p(x) — \\x\\i • &{w?r) if
x / 0, where || • ||x denotes the standard norm of ?\. Then ip is continuous,
one-to-one, and maps ?\ onto ?p. To see the latter, given y ? lp, y ^ 0, let
x = (sign(y2-)g). Then ^(^||y||P) = ||y||p • ^ = y.

10. Smoothness and Structure 337
We claim that tp l is continuous; that is, <p is a homeomorphism. Let
yk 6 ?P, yk -> 0, yk ^ 0. Let z* G S^ be such that <p(xk) = j^-. Then
p(lly*llp**) = Vk a*d ||2/*||Ps* -> 0, so tp-\yk) - 0.
Now take yfc E ?p, y € ?p such that yk -^ y, y ^ 0. Let xfc G S^ be
such that <p{xk) — ir^np, and let x G S^ be such that cp(x) = \\?\T- Since
Ih/Ml *" IMF' we &e^ ^ ~~* ^ because (p — <p is a homeomorphism of S^
onto 5/p. We have p(||2/%z*) = yk,<p(\\y\\px) = y, and ||t/*||j,:e* -+ ||2/||Pz.
Therefore <p~l{yk) —> <?>_1B/).
D
Torunczyk proved by topological methods that all infinite-dimensional
Banach spaces of the same density character (see Chapter 11) are mutually
homeomorphic ([Tor]). We will prove here an earlier result that is a special
case of Torunczyk's result (see also [Bes3]).
Theorem 10.35 (Kadec [Kad3])
All separable infinite-dimensional Banach spaces are mutually
homeomorphic.
We will prove it for the case of reflexive spaces and refer to [BeP2] for
the general case.
First, we will do some preparatory work. Let X be a separable reflexive
infinite-dimensional Banach space. By Theorem 8.17, we may assume that
the norm of X is locally uniformly rotund. Let {z2; /,•} be a Markushevich
basis of X (Theorem 6.41). For n G N0, put Xn — span{zn+i,...}. Note
oo
that p| Xn — {0}. Indeed, given i G N, we have fi(x) — 0 for every
n = l
x G Xi and thus fi(x) — 0 for every x ? f)Xn and every i. Since {/;}
separates points of X, we get that f]Xn = {0}. Also, Xn + span(#n) is
closed in Xn_i and Xn + span(#n) contains {xn,a:n+i,...}; hence Xn +
span(xn) = Xn_i. Thus, Xn is a hyperplane in Xn_i generated by fn that
splits Xn_i into three mutually disjoint parts: Xn and the open halfspaces
^-1 = /n-HO.Oo), X~_, = f-H-00,0).
By induction, the codimension of Xn in X is n. Note that the distance
function to each Xn is weakly continuous. Indeed, if zk -^> z and zki z
denote their cosets in X/Xn, then zk ^ z in X/Xn. Since X/Xn is finite-
dimensional, Zk —> z in X/Xn and thus distB:jb,Xn) = \\zk\\ —* ||^|| =
dist(z, Xn). Note that, given n and x G X, there is a unique zn G Xn such
that \\x — zn\\ = dist(z,Xn) (Exercise 8.36).
Finally, for every x G X, we have 0 < dist(#,Xn) < dist(x, Xn+\) < \\x\\
and lim (dist(x,Xn)) = ||#||. To see the latter, let zn G Xn be such that
n—>-co
\\x — zn\\ — dist(#,Xn) for every n. Then zn —> 0. Indeed, given i G N,
fi(zn) = 0 for n > i because /t- = 0 on Xn. Since Bx is weakly compact
and the topology on Bx of the pointwise convergence on {/;} is Hausdorff,

338 10. Smoothness and Structure
the weak and the {/2}-topologies coincide on Bx> Because {zn} is bounded,
From the weak lower semicontinuity of the norm, we thus get ||ar|| <
liminf ||x — zn||. Therefore ||a?|| < liminf \\x — zn\\ < limsup ||a? — zn\\ < \\x\\
n-*oo n-+oo n-*oo
as \\x — zn\\ = dist(#,Xn) < \\x\\ for all n. Hence lim (dist(#,Xn)) = \\x\\.
n—>oo
For n E N and x E X, define Hn(x) = dist(#,Xn) = \\x — zn\\ for some
unique zn E Xn. Furthermore, to every x ? X we assign a sequence {hn(x)}
defined inductively by
hrx)-i #i(*) ifxGX+UXi
^lW~\ -Hi(x) ifxeX'
( Hn+1(x) ifzneX+
hn+i(x) = < hn{x) if zn E Xn+i (i.e. if zn+1 = zn)
[ -ff„+i(x) iiznex~.
We now have the following:
Proposition 10.36
In the above notation, lei {hn} be a sequence of real numbers such that
\hn+i\ > \hn\ for every n, sup |/in| = 1, and An+1 = hn if \hn+i\ = \hn\.
Then there is a unique x E Sx such that hn(x) = hn for every n.
The proof of Proposition 10.36 is based on the following lemma.
Lemma 10.37
Lei hi,...,hn be a sequence of real numbers such that |A«+i| > \hi\ for
every i < n and hi+i = hi if |At+i| = l^l- Let Qn be the set of all x E X
such that hi(x) — hi fori = 1,..., n. Then Qn — an+Xn for some an E X.
Proof: For n = 1, the set Qi is a shift of X\ and the result holds. Assume
the assertion holds for hi,..., /in_i and denote the resulting affine subspace
by Qn-i- The set Qn is obviously a subset of Qn-\- Let x be an element
of Qn-i with minimal norm, in particular ||ar|| = |/*n-i|, and let y be an
arbitrary element in X*_ x.
Given A E R, for every element x\ of the affine subspace Qn,x ~ Xn +
x + Ay we have Hi(x\) — \h{\ for i = 1,..., n - 1, since Qn^\ C Qn-i
and Hn(x\) = V>(A). If we vary A, the function t/>(A) is a strictly convex
function that attains its minimum at 0 with ip@) = |/*n-i|- The equation
ip(X) = \hn\ for \hn\ > |An_i| has exactly one positive and one negative
solution.
Each of these solutions gives us one of the affine subspaces Qn,o, Qn,Ai,
and Qn,A23 whose elements satisfy hn(x) = /in_i(x), ftn(^Ai) = |^n|, and
hn(x\2) — — \hn\, respectively, by the definition of hn(x). This completes
the proof of Lemma 10.37.
?

10. Smoothness and Structure 339
Proof of Proposition 10.36: The set 5 of all elements x E X such that
hn(x) — hn for every n is the intersection of the sets Qn from Lemma 10.37.
Since Qn is a shift of Xn, if the intersection of all Qn contained at least two
different points, then shifting back to n we would have that f]Xn would
contain at least two different points, which is not the case. Therefore, the
set S consists at most of a singleton. To see that S is nonempty, consider
the intersection of Qn with Sx- Since \hn\ < 1 and the distance of Qn to
Xn is |/in|, we have Sx H Qn ^ 0 (here we used the refiexivity of X). The
weak compact sets Qn 0 Bx form a nested family of weak compact sets,
and thus their intersection is nonempty and contained in Sx-
?
Proof of Theorem 10.35: Let X) Y be infinite-dimensional separable
reflexive Banach spaces. We assume that their norms are locally uniformly
rotund and that {xi]fi] and {yi\gi} are Markushevich bases of X and Y,
respectively. We construct closed subspaces Xn of X and Yn of Y as above,
and define H*,h* on X and H% ,h% on Y accordingly. We construct a
map (p of Sx into Sy as follows. Given x E Sx> let <p(x) be the unique
element of Sy such that h%(cp(x)) = h* (x) for all n. Then <p is one-to-
one and onto Sy by Proposition 10.36. Since the roles of X and Y are
interchangeable, to prove that ip is a homeomorphism of Sx onto Sy we
need only show that <p is continuous.
To this end, assume that xk,x E Sx, and xj. —> x. Since y>(a?fc),^(a:) E
Sy and since on Sy the norm and weak topologies coincide, it suffices to
show that (p(xk) —> <p(x). Assume the contrary. Then (using refiexivity and
separability), {<?>(#*:,)} -+ y € By for some subsequence {&/} of N, and
y ^ p(z). Let n0 = min{n E N; /?(y) ^ A* (a?)}. Then either |/#0(y)| >
ltf„(*)l > K-lMI. °r l&n„(*)l > l^o-lWI-
Assume without loss of generality that \h%o(y)\ > |/i^o_1(y)|. We know
that /z*o is continuous in the weak topology and thus in the {/;}
topology at points p E Bx where \h^0_1(p)\ < \h*0(p)\. This is due to the
fact that sign(/i^o) is continuous in the {/«}-topology at points where
xno-i ? Xno. Consequently, lim h%o((p(xkl)) = h%Q(y). By the assumption,
however, lim \h% (<p(xkl))\ = lim \h* (xkl)\ = |^0(z)|, a contradiction.
Therefore, <?> is continuous and <p is a homeomorphism of Sx and Sy. The
homeomorphism of X onto y can then be constructed as in the proof of
Proposition 10.34.
?
The situation with weak topologies is less clear. For instance, it is not
known whether for James's space J, (J, w) is homeomorphic to (J0j,iy).

340 10. Smoothness and Structure
Smoothness in ?p
Theorem 10.38 (Meshkov; see, e.g., [DGZ3])
Let X be a Banach space. If both X and X* admit a F2-smooth bump, then
X is isomorphic to a Hilbert space.
Proof: ([FaZ2]) First, note that X is an Asplund space (Corollary 10.9).
Let f.X -» R, g:X* —» R be F2-smooth bumps. Assume that /@) ^ 0
and put ip = f~2 (define ip(x) = +oo if f(x) — 0). Then the conjugate
function rp* is convex and locally bounded, and hence continuous on X*
(Exercise 8.1). Applying Theorem 10.20 to Z — X and <p = —-0* + g~2, we
obtain x*0 G X* such that -V>*(*5 + h*) + 9~2{*o + h*) ~ ^(xo ~ h*) +
g~2(x*0 - h*) + 2^*(^S) - 2flf-2(a?5) > 0 for all A* G X*. Hence g{xl) ? 0,
and for all h G X*,
^*(xS+A*)+^*(xS-A*)-2^(x;) < y-2K+ft*)+^2D-/i*)-2^-2D).
The function g~2 is F2-smooth at xjj, so there are c, 5 > 0 such that
g~2(xl + ft*) + g~2(xl - &*) - 2^K) < c#*||2
for all /i* G X*, ||A*|| < E. Hence V*K + h*) + ^*(*o - h*) - 2t/>*(x*0 <
c\\h*\\2.
Because ip* is convex and continuous, the last inequality implies that $*
is Frechet smooth at a?j$. Denote x0 = iP*'(xq). Then a(^*, a?o, ^0,0 < c/2
for 0 < t < 6 in the notation introduced prior to Lemma 10.17. Using
Lemma 10.18, we get x0 G X, and by Lemma 10.17, ct2+f3(ip**, xq, Xq, s) >
ts for all 0 < t < 6. Hence, using Lemma 10.18, C(tp,xo,XQ,s) > j^s2 for
all 0 < s < 2c6. Thus,
W*o + h)-1>(zo)-xi(h)>±\\h\\2
for all h G X, \\h\\ < 2c6. Since ^ is F2-smooth at xq) by l'Hopital's rule
we get il)n{xQ){h) h) > j^\\h\\2 for all h G X. Because i/>"(xo) is a bounded
bilinear form on X, we conclude that h i—> [ip/f(xo)(h, h)] is an equivalent
Hilbertian norm on X.
' ' ?
Definition 10.39
Let f: X —> (—oo, -j-oo] 6e a proper function on a Banach space X and
x G Dom(/). Let p G [l,oo). We say that f is Tp-smooth at x if there is a
polynomial P: X —» R of degree at most p such that
f(x + h) = f(x) + P(h) + o(\\h\\*) for all heX.
Let Q be an open set in X. We say that (p is Tp-smooth on Cl if it is
Tp-smooth at every xEll-

10. Smoothness and Structure 341
Lemma 10.40
Let (p: X —> (—00,-foo] be a proper function on a Banach space X and
x G Dom(^); p G N. If (p is Fp-smooth at x, then (p is Tp-smooth at x.
Proof: For h G X, denote
P(h) = <p'(x)(h) + fy"(x)(h, A) + • ¦ ¦ + ±<p<*\x)% ..., h).
Applying l'Hopital's rule p — 1 times, we get
\^(>p(x + th)-cp(x)-P(th))
= lim -J— U*-l\z + th)(h, ...,/>)- V{p-l\x){h, ...,h)
t—>0 p • • • Zt \
-^(*)(/*,...,/M/0) =0.
Noting that all these limits are uniform with respect to h G Bx> we can
conclude that (p is Tp-smooth at x.
?
Lemma 10.41
Let p G (l,oo), and let {ei}fl1 be the canonical basis of?p.
(i) IfP: X —> R is a polynomial of degree less than p, then lim (P(e2)) = 0.
(ii) If x G ?p, x ^ 0; then fort G @, \\x\\p) we have
lim Wz + taW, = (||*||?+ <"I/P > ||*||P + ^lkl|^pfp.
Proof: (ii): The equality is proven in a standard way, first considering x
with a finite support. The inequality follows from the strict concavity of
the function t^t1^, t > 0.
(i): Since e2- ^ 0, P(ei) —* 0 for every polynomial of degree 1. Take
any integer 1 < n < p and assume we have already verified (ii) for all
polynomials of degree less than n. Let P be a polynomial of degree n.
By contradiction, assume that limsup |P(ez)| = 3a > 0. Without loss of
2—*- CO
generality, we will assume that P(e2) > 2a for all i G N. We observe
that P(x + h) = P(x) + Q(x, h) + P(h) for x,h ? ?p, where Q(z, •) is a
polynomial of degree less than n. Put #i — e\. If a?a- has been constructed,
using the induction assumption, find j G N so that Q(xiJej) > —a. Put
#z+i = a?j + e;-. Then
P(xi+1) = P(ar,-) + Q{xi,ej) -f P(ej) > P(^) + a > . . . > (i + 2)a.

342 10. Smoothness and Structure
Note that ||a?t-||p = i1^ for all i G N. Therefore,
p& >^- = i^'Pa -ooasi-oo,
Ikill? in/p
contradicting the fact that {'LA,^'; \\h\\ > 1} is bounded.
D
We note that there is a reflexive separable infinite-dimensional Banach
space X such that all polynomials on X are weakly sequentially continuous
(the Tsirelson space).
Theorem 10.42 (Kurzweil; see, e.g., [DGZ3])
Let p G A, oo). Ifp is not an even integer, then the space ?p does not admit
any continuous Tp -smooth bump.
Proof: ([FaZ2]) By contradiction, assume that b:?p —» R is a Tp-smooth
continuous bump. Applying Theorem 10.20 to (p — b~2 — || • || (and to
Z* =tp* =?q,q= ^y), we get x e ?p such that b~2(x+ h) + b~2(x - h) -
2b~2(x) > \\x + h\\p + \\x-h\\p-2\\x\\p for all h G ?p. Then b(x) ^ 0 and we
check that b~2 is Tp-smooth at x. By the above inequality and definition
of Tp-smoothness, for all h G ?p we have
P(h) + o(\\h\r)>\\x + h\\p + \\x-h\\p-2\\x\\p,
where P is a polynomial of even degree, say n, with n < p. (If P is a
polynomial of odd degree, then h i—> P(h) + P(—h) is a polynomial of
smaller and even degree satisfying the above property.) Fix any t > 0 and
consider the last inequality with h — te{, i G N. Since p is not an even
integer, then n < p and Lemma 10.41 (ii) yields
o(tp) = hmm{(P{tei) + o(\\tei\\p)) > lim (||ar + /et-||J, + ||a:-*cl-||p-2||xy.
i—+oo i-+oo
Now, if x = 0, then o(tp) > 2/, a contradiction. If x ^ 0, then for 0 < t <
\\x\\p we have by Lemma 10.41 (i) that o(tp) > ^\\x\\p~ptp, which is again
a contradiction.
D
Theorem 10.43
Let p G (l,oo). Ifp is even, then the canonical norm of ?p is C°°-smooth.
If p is odd, then the canonical norm of ?p is C1?'1-smooth.
Ifp is not an integer, then the canonical norm of ?p is C^-smooth, where
[p] is the integer part of p.
We refer to [DGZ3] for a standard and lengthy proof.
The best order of Gateaux smoothness for an equivalent renorming of ?p
is not known.

10. Smoothness and Structure 343
Countable James Boundary
A typical example of a Banach space with a countable James boundary is
Co or C(K) for some countable compact K.
Theorem 10.44 (Fonf, Kadec; see, e.g., [FLP])
Let X be an infinite-dimensional Banach space. If X has a countable
James boundary, then X is saturated with subspaces isomorphic to c0;
that is, every infinite-dimensional closed subspace of X contains a subspace
isomorphic to cq.
In the proof, we will use the following notion.
Definition 10.45
We say that a real-valued function (p on a Banach space X locally depends
on finitely many functionals (or coordinates) if for every x G X there is a
neighborhood U of x, /i,. .., fn G X* and a continuous real-valued function
i\) on Rn such that <p(z) = i>(h(z)> ^(z),. ¦ •, fn{z)) for all z eU.
The supremum norm in cq is an example of a function that locally
depends on finitely many coordinates away from the origin.
Proof of Theorem 10.44: Observe that the restrictions of the elements
of a James boundary to a closed subspace form a James boundary of
that subspace. Therefore, it suffices to show that X contains a subspace
isomorphic to cq.
Assume that this is not true. Let B — {y*} be a countable boundary of
X. Define a norm ||| • ||| on X by
|*| = sup{(l+ !)!/„(*); neN}.
If x G X is such that \\x\\ — 1 (|| • || denotes the original norm of X), then
there is n such that 2/*(#) — 1 and thus |x| > (l + ^)y^(x) > 1. On the
other hand, if ||x|| = 1, then (l + ?)i?(s) < (l + ?)||^|||k|| < 2 for every
n. Hence ||| • ||| is an equivalent norm on X. We will show that, away from
the origin, the norm ||| • ||| locally depends on finitely many {y* }. Let n0 G N
be such that Vn0(x) = 1- For n > n0, we have
(l + ?)!?(*)<l + ?<l + ;dT<l + ?-
Hence
sup{(l + $)y'n(x); n > n0} < 1 + ^ < 1 + ?.
On the other hand,
sup{(l + IJy^ar); n < n0} > (l + ^^.(x) = 1 + ?.
Since the supremum of a family of 2-Lipschitz functions is a 2-Lipschitz
function, from the last two inequalities we deduce that there is a

344 10. Smoothness and Structure
< 1 for all k G N.
neighborhood U of x such that, for all z G f7,
^{(^^nWi w<no}>sup{(l + i)y;(z); n>n0}.
Therefore, |z| = sup{(l + ^)Vn{z)\ n < no} for all z G 17, showing that,
away from the origin, the norm ||| • | locally depends on finitely many {y*}.
Let r be a continuous real-valued function on the real line such that
t(x) — 1 for x G (— |, |) and r(z) — 0 if \x\ > 1. Define a function on X
by g(x) = 1 — r(|ar|). Then # is continuous and locally depends on finitely
many {?/*}. Note that g@) = 0 and g(x) = 1 for |x|| > 1.
We will construct inductively a sequence {xn}^L0 Clas follows: set
xq = 0 and if #o, a?i,... ,arn have been chosen, choose xn+i so that:
(!) ^(E^^ + ^n+i^n+i) = 0 fore,- = ±1, i = l,...,n+l;
B)||a?n+1||>iMn==|sup||y||J
where the supremum is taken over all y — xn+i satisfying A). Since
(\ ii i
Yleixi) = 0 for every k G N, we get MC^d
2=1 ' »i=l _ '
Since we assumed that X does not contain an isomorphic copy of Co,
by Theorem 6.38 we have that Ylxi *s unconditionally convergent and
5 = < ^2 E{Xi\ ei = ±1, n G N > is relatively compact in X (Exercise 1.38).
n=i ^ _
By the local dependence of g on finitely many {y?}, for each z G S there is
<5Z > 0 and a finite set Kz of natural numbers such that g{w ± 6x) = #(iu)
for all tu G B(z,6z), 6 < SZi and x ? f] (y*J_1@). By compactness,
_ *
S C |J B(zi,6Zi) for some z2- G X. Let 5 = min{<$2.; i — l,...,fc} and
2 = 1
k
K = ij #,.. Let x G fl (l/*).7 (°)> ° < IWI < 6. It w e S, then w G
i=l «?l<:
B(ziif>zi) for some ? and hence g(w ± <5z) = #(w). Thus, inf(Mn) > <5 > 0
for all n G N, which contradicts the convergence of J2xn-
n
Theorem 10.46 ([Hajl])
Let X be a Banach space. If X has a countable James boundary, then X
admits an equivalent C°°-smooth norm that locally depends on finitely many
coordinates.
PROOF: (Sketch) We will only prove the existence of a C°°-smooth
equivalent norm. Bx and B% (the open unit ball) refer to the original norm || • ||
of X. Let B = {y*} be a James boundary of X. Define a function ip on B%
by
^) = ?)(l + i)y;(x)a
\2n
n = l

10. Smoothness and Structure 345
The definition is consistent, because <p(x) < 2 Yl \\x\\2n — l-iUiu- The
n = l
function <p is C°°-smooth, since it is in fact given by its Taylor series
converging absolutely and uniformly on every AJ3x, A < 1.
Let S = {x\ (p(x) < 1}. Then S is symmetric, convex, and closed, and
contains a neighborhood of the origin. By the implicit function theorem
([Die]), the Minkowski functional of S is a C°°-smooth equivalent norm
onl.
?
Corollary 10.47 (Haydon [Hayl])
Let K be a countable compact. Then C(K) admits an equivalent C°°-smooth
norm.
Proof: C(K) has a countable James boundary B = {±<5jt; k ? K}, where
8k are Dirac functionals, so we can apply Theorem 10.46.
?
Theorem 10.48 ([DGZ3], [FaZl])
Let X be a separable Banach space. If X admits a continuous bump (p that
locally depends on finitely many coordinates, then X* is separable and X
is saturated with isomorphic copies of cq.
PROOF: The statement on the saturation follows from the proof of
Theorem 10.44. We will prove that X* is separable. For every x ? X, we choose
a neighborhood Ux in X, functionals /f,..., /? ? X*, and a continuous
function ipx on Rnx such that, for every z EUX,
By the Lindelof property of X, let {?n}?°=1 C X be such that {UXn}(^-1
oo
covers X. For n ? N, denote Fn = {/fn,..., f%» } and F = (J Fn. Then
n — l
tp is a continuous function on X with bounded nonempty support, cp locally
depends on finitely many elements of F} and F is countable.
Define a function $ on X by
*W-\ +oo if^(ar) = 0.
Then $ is a bounded-below lower semicontinuous function on X such that
5 = {x ? X; <?(#) < oo} is open, and on its domain $ locally depends on
finitely many elements of F.
Because F is countable, in order to show that X* is separable, it suffices
to prove that span(F) = X*. Fix any / ? X* and e > 0. By Theorem 3.53,
there is x$ ? S such that
(*-/)(*)>(*-/)(*o)-e||*-*o|| (*)

346 10. Smoothness and Structure
for all x G X. Let U, /i,..., /n G F, and t/> be as in the definition of the local
dependence for $ — / at xq. Set W = {x ? X; fi(x) = ... = /„(#) = 0}.
Finally, let 6 > 0 be such that x0 + h G U whenever h e X, \\h\\ < 6. Then
ifheW, \\h\\ < S, we have
<&(X0 + h)- $(*0) = *l>(fl(x0 + A), . . . , /n(^0 + A))
-^(/i(^o),..-,/n(a?o))
= ^(/l(*o), . • ¦ , /n(*o)) - V>(./lOo), . . . , /n(«o))
= 0.
Hence from (*), for h G FT, ||/i|| < S, we have
/(A) = f(x0 + h)- f(x0) < $(x0 + h)- $(x0) + e||/i|| - e\\h\\.
Let // = | Then, by the last inequality, for / as an element of W*
we have ||/|| < e. Let g be a norm preserving Hahn-Banach extension
of / to X. Note that / — g G WL. Since span{/i,..., fn} is u>*-closed
in X* as a finite-dimensional subspace, by the bipolar theorem WL —
span{/i, ...,/„}. Hence
dist(/,span(F)) < dist(/,span{/i,..., /„}) = dist(/, W1)
< \\f-{f-9)\\ = \\9\\<e-
Thus span(F) = X*.
?
Exercises
10.1 Let X,Y be Banach spaces, and let h be a Gateaux (Frechet) differ-
entiable homeomorphism of X into Y. Show that if Y admits a Gateaux
(Frechet) smooth bump, then so does X.
Hint: Use the composition of maps and the chain rule for differentiation.
10.2 Assume that a Banach space X admits a Gateaux (Frechet) dif-
ferentiable bump 6. Show that X admits a bounded Gateaux (Frechet)
differentiable bump.
It is not known whether X admits a Lipschitz, Frechet differentiable
bump if X admits a Frechet differentiable bump.
Hint: Assume that 6@) = 1 and b(x) = 0 for ||z|| > 1. Put xj){x) = l-e^)
for x G X.
10.3 Use Theorem 10.6 to show that ^ does not admit any continuous
Gateaux differentiable bump function.

10. Smoothness and Structure 347
Hint: dens(^) > card(/?N) = 2C > cardDo) = c.
10.4 Show that t^ admits no equivalent Gateaux differentiable norm.
Hint: The norm ||| • ||| defined on t^ by |z| — ||^||oo -flimsup \x{\ is nowhere
i—>-oo
Gateaux differentiable (Exercise 8.32). If t^ admitted a Lipschitz, Gateaux
differentiable bump function, then by Corollary 10.9 we would have that
I • I must be Gateaux differentiable on a dense set of points.
10.5 Show that ?i(T) admits no Lipschitz, Gateaux differentiable bump
function if T is uncountable.
Hint: The standard norm of ?i(T) is nowhere Gateaux differentiable
(Exercise 8.26). Then use the same argument as in the previous exercise.
10.6 Let (X, || • ||) be a Banach space. Define a function / on X by f(x) =
\\x\\. Calculate the conjugate function /* to /.
Hint:
f*(x*) - sup {?*(#) - ||s||} = sup{ sup {x*(x) - ||z||}}
x?X r>0 \\x\\=r
= sup{r||x*|| — r} = sup{r(||?*|| — 1)} = +oo
r>0 r>0
if ||x*|| > 1 and /*(**) = 0 if ||x*|| < 1.
10.7 Let (X, || • ||) be a Banach space. Define a function / on X by f{x) — 0
for x G Bx and f(x) = -foo otherwise. Calculate the conjugate function
r to /.
Hint:/* = 11-11*.
10.8 Let (X, || • ||) be a Banach space. Define a function / on X by f(x) =
|||x||2. Calculate its conjugate /* on X*.
Hint: Note that f*(x*) = sup{sup{#*(#) - /(#); \\x\\ — r}}. Use the defi-
r>0
nition of ||#*|| to calculate the "inside" supremum and then use elementary
calculus to obtain the final result.
10.9 Use Theorem 10.20 to prove Pitt's theorem: Given 1 < p < q < oo,
then every T G B(?q}?p) is a compact operator.
Hint: Let {x{} be a bounded sequence in ?q. We may assume that xi -^ y\
then T{yi) ^ T(y). Apply Theorem 10.20 to x h+ ||z||| - ||T(z)||?- Then
for h = t(xi — y), t > 0, i G N, we get
\\x + t(xi-y)\\\+\\x-t(Xi-y)\\l-2\\x\\\
> \\T(x)+tT{Xi - y)\\> + \\T(x) - tT(xt - y)\\r - 2||T(z)||?,
and 2ti limsup^^ ||ar,—j/||| > W limsup,-^ ||T(a;,—y)% for t > 0. Hence
\\T(xi - y)\\p —> 0 as i —> oo.

348 10. Smoothness and Structure
10.10 Assume that the norm || • || of a separable Banach space X can
be approximated by a Lipschitz C1-smooth function (p on X such that
|||x|| — <p(x)\ < 1 for every x G X. Is X* necessarily separable?
Hint: Yes. By considering a C1-smooth function r on R such that r(t) — 2
whenever t G [-1,1] and r(t) = 0 for \t\ > 4, we have r(^@)) — 2,
r((p(x)) — 0 for ||x|| > 5, and r o cp is C1-smooth. Use Corollary 10.9.
10.11 Assume that <p is a nonnegative C^-smooth bump on Rn such that
/Rn (f(t) dt — 1 and supp(<?) C 6B-&n.
Show that / is a uniformly continuous convex function on Rn, then
f^x) = JR7l f(x — t)cp(t) dt provides a good convex smooth approximation
of / in the sense that \\f^ — f\\c(Kn) —> 0 as <5 —> 0.
This method cannot be used in infinite dimensions.
Hint: Estimate \f(x) - f6(x)\ < fKn \f(x) - f(x - t)\<p(t) dt.
10.12 ([Cep]) Let / be a Lipschitz function on a uniformly convex Banach
space X. Then there is a sequence {/n} of A-convex functions (i.e.,
differences of convex functions) that are bounded on bounded sets of X and
converge to / uniformly on bounded sets.
Hint: Assume that the Lipschitz constant of / is 1. For n G N and x G X,
define fn(x) = inf {f(y) + 2n||^||2 + 2n\\y\\2 - n\\x + y\\2}.
Then fn = cn — dn, where cn = 2n||z||2 and dn(x) — sup{n||x -f y\\2 —
y?X
2n\\y\\ — /(?/)}• The functions cn and dn are convex and, by taking y — x
in the infimum, we can see that fn < f for all n. Since 2||x||2 + 2\\y\\2 —
\\x + 2/||2 > 2||z||2 + 2\\y\\2 - (||x|| + ||x/||J = (||x|| - ||»||J > 0, /„ is an
increasing sequence of functions.
Let y be such that f(y) + 2n||z||2 + 2n||y||2 - n\\x + y\\2 < f(x). Then
since / is 1-Lipschitz,
"A1*11 - IbllJ < 2n||x||2 + 2n\\y\\2 - n\\x + y\\2 < f(x) - f(y) < \\x - y\\.
Suppose that ||y|| > 1 + ||a;||. Then
1 < |INI - lbll| < (INI - INIJ < ?||* - 2/11 < ?||*|| + ?||y||.
Hence, for n > 3 we have ||y|| < ^y||^|| < 2||je||. Thus, in any case we have
\\y\\ < 2A + ||x||). This implies that
fnW = n u^L in{/(y) + 2n|H|2 + 2n||y||2-ri||^ + y||2}.
I|y||<2(i+IMI)
Hence, in particular, fn is bounded on bounded sets and so is dn = cn — fn.
Moreover, for such y, we have
0 < 2\\x\\2 + 2\\yf - \\x + y\\2 < ±||* - y|| < I||*|| + ±\\y\\ < f A + ||*||).
Assume that it is not true that fn—*f uniformly on bounded sets. Then
there is e > 0 and a bounded sequence {xn} such that fn{xn) + e < f(xn)

10. Smoothness and Structure 349
for all n ? N. For each n, choose yn such that
fn(Vn) +e< f(yn) + 2n||zn||2 + 2n\\yn\\2 - n||xn + yn\\2 + e < f(xn).
Then we have \\xn - yn\\ > f(xn) - f(yn) > ? and
0 < 2||*n||2 + 2\\yn\\2 - \\xn + yn\\2 < f A + ||xn||) - 0
as n —> oo. Thus X is not uniformly convex, a contradiction.
10.13 Show that every uniformly continuous map / of a Banach space X
into a Banach space Y is Lipschitz for large distances; that is, for every
8 > 0 there is K = #(«) such that ||/(ar) - f(y)\\ < K\\x - y\\ for all
x, y ? X satisfying \\x — y\\ > 8.
Hint: Given 8 > 0, first get M such that \\f(a) - f(b)\\ < M for a, 6 ? X
satisfying \\a — b\\ < 8 as follows. Using the uniform continuity for e = 1,
find A > 0 such that x,yeX, \\x - y|| < A implies \\f(x) - f(y)\\ < 1. If
\\a — b\\ < 6, we can get from a to 6 in [6/A] steps ([w] denotes the integer
part of w) not longer than A, so ||/(a) - /F)|| < 8/A — M.
Put K — 2M/6. Given x, y ? X with \\x—y\\ > 6, let x — ao, a\)..., am =
y be points on the line segment [x,y] such that ||aj+i — aj\\ < 8 and
m= [2||x-y||/«]. Then
m
ll/(*)-/(y)ll < X)H/(a«-)-M--i)ll<"»^
t = l
< taJl/^^A^.^i.
10.14 Let X, y be Banach spaces. Assume that Sx is Lipschitz equivalent
to Sy via a map (p. Show that X, Y are Lipschitz equivalent.
Hint:zHMb(fl%)-
10.15 ([BeLi]) Define r(t) = 2* for * > 0 and r(t) = * for t < 0. Define
yr.^2 —> ^2 by ^((a?2)) = (r(#z)). Show that 9? is a Lipschitz map from ?2
onto ?2 that is nowhere Frechet differentiable.
Hint: Direct calculation.
10.16 Let (p be a Lipschitz homeomorphism of X onto Y that has a Frechet
derivative at some point x$. Show that ^'(^o) is a linear isomorphism of
X onto Y.
Hint: Assume that <p and <^_1 are if-Lipschitz, xq = 0, and <?>@) = 0. Let
T = y>'@). Find 8 > 0 so that ||p(A) - T(A)|| < |||A|| whenever \\h\\ < 8.
From the proof of Theorem 10.33 we get that T is an isomorphism of X
into y. Assume T(X) / Y. Since T(X) is closed, find y ? Y such that
||t/|| < ? and dist(y,T(X)) > f. Let x = ^(ar). Then ||x|| < 8, so we get
a contradiction:
f < dist(y, T(X)) < \\y - T(x)\\ = \\<p{x) - T(x)\\ < §.

350 10. Smoothness and Structure
10.17 Let P be a Lipschitz (nonlinear) projection of a Banach space X
onto a subspace Y of X such that it is Gateaux differentiable at a point of
Y. Then Y is complemented in X.
Hint: The derivative is a linear projection.
10.18 Show that if Y is a reflexive subspace of a Banach space X and there
is a uniformly continuous nonlinear projection P of X onto Y, then there
is a Lipschitz nonlinear projection of X onto Y.
Note that for every closed subspace of an arbitrary Banach space there
is a continuous nonlinear projection onto it. This is a consequence of
Dugundji's theorem (see [Dugl]).
Hint: Since P is Lipschitz on large distances (Exercise 10.13), there is A > 0
such that \\P(x) - P(y)\\ < X\\x - y\\ for all x,y E X with \\x - y\\ > 1.
Consider projections Pn of X onto Y defined by Pn(x) = ^P(nx) for
n E N. Note that, for a fixed xGl,
\\Pn(x)\\ = \\Pn(x) - P„@)|| - ||lp(nx) - iP@)|| < iA|M| = A|H|
for \\nx\\ > 1 (i.e., for n > \\x\l~1). Since Y is reflexive, By is weakly
compact and thus, by TychonofTs theorem, {Pn} has a cluster point Po
in the topology of the pointwise convergence for maps into Y in its weak
topology.
Because Pn{y) = t/ for all y E Y, we have Po(y) — y for all y E Y.
If x,y E X, then \\Pn(x) - Pn(y)\\ < X\\x - y\\ for all n > \\x - y||_1.
Therefore, ||Po(^) — -Po(y)|| < M\x — y\\ for all x,y ? X. Hence Po is a
Lipschitz projection of X onto Y.
Note that this is the first step in the proof of Lindenstrauss's result (see,
e.g., [BeLi]) that if Y is a reflexive subspace of a Banach space X and there
is a uniformly continuous nonlinear projection P from X onto Yy then Y
is complemented in X.
10.19 There is a separable Banach space X that is not isomorphic to a
Hilbert space and has a closed subspace Y such that both Y and X/Y are
isomorphic to Hilbert spaces ([ELP]). Show that for this space there is no
Lipschitz selector i\)\ X/Y —> X with the property 7r(ip(x)) = x for every
x(EX/Y.
Hint: If such ip existed, by the proof of Theorem 12.58, X would be Lipschitz
equivalent to Y 0 X/Y, which is isomorphic to a Hilbert space. Then X
would be isomorphic to a Hilbert space by Corollary 10.32.
10.20 Let X, y be Banach spaces and T E B(X, Y), T is onto. The Bartle-
Graves theorem asserts that there is a continuous (in general, nonlinear)
map cp from Y into X such that T((p(y)) = y for every y E Y (see, e.g.,
[DGZ3]). Show that if Ker(T) is not complemented in X, then ip is nowhere
Gateaux differentiable.

10. Smoothness and Structure 351
By the differentiability results for Lipschitz maps, it therefore follows
that, in general, there is little hope of having Lipschitz selectors of the
Bar tie-Graves type (see [BeLi]). If X is uniformly convex, then the
existence of the Bar tie-Graves map <j> that is uniformly continuous on bounded
sets is discussed in [BeLi].
Hint: Otherwise, Ix — <pf(y)T is a projection of X onto Ker(T).
10.21 Let X be a finite-dimensional Banach space. Show that Bx is not
homeomorphic to Sx •
Hint: Let <p\ Bx —> Sx be a homeomorphism. Define t{x) — —x on Sx and
consider the Brouwer fixed point theorem for <p~l o r o (p\ Bx —> Bx-
10.22 Let X be a Banach space. Show that h(x) — lJ\\x\\ defines a
homeomorphism of B^ onto X.
10.23 Let X, Y be Banach spaces. Show that if (p is a homeomorphism of
Sx onto 5y, then x i-> ll^llvKjifyr)' 0 h-> 0, is a homeomorphism of X onto
y.
10.24 Let Si and 52 be spheres of two equivalent norms on an X. Show
that the radial map x i-» jr4p is a Lipschitz homeomorphism of Si onto
52- Is it a homeomorphism in the weak topologies?
Hint: About the ^-topology: No. Try a norm with the Kadec-Klee property.
10.25 Show that Stp is uniformly homeomorphic to Si1 for every p E
[1,00).
The space lp is not uniformly homeomorphic to ?q if p / q ([BeLi]),
although the extension of the map <p to the balls by ||(aOllvKn^Hir) S*ves
a uniform homeomorphism of Bip and Biq.
Hint: For simplicity, assume p — 2. Define a bijection <p of Si2 onto 5^ by
^((a*)) = (a? sign (a,-)). If (a«)»F0 G &2> tnen
lM(«0)-?((*0)ll = J]K2sign(a0-^2signF0l
= Yl ia>2 sign (a<)" blsign Fz*)i
+ ]T |az2sign(a2) - 6?signF2-)| = DA + DB,
B
where A={nE N; sign (a;) / signF?)} and B — {n E N; sign(a2-) =
sign F.)}. Using ||^|l| < 2||ar||2 for ||x||2 < 2, we get
Da = ?a? + &H?H2 + M2<DH + MJ
= ?|a,-it|2<^|a,-lIf)i,
A A

352 10. Smoothness and Structure
and
Db = ]P fa - bi\ \cii + b{\ < Y^ \ai ~ b*\ \a*\ + Y2 \ai ~ bi\ l6z'l
B B B
< (Ei*-*«i3)*(EN2)* + (i;^-Ma)*(EN2)*
B B B B
< 2(^k--m2)*.
The lower estimate follows from ^ a?+6? > ^ X^(a*""^'J anc^ S lai """"&? I >
,4 a ?
B
10.26 (Gorin) Show that for every n, ?Vj, ls uniformly equivalent to a
bounded subset of ?2] that is, there is a map <p from ?% onto a bounded
subset of ?2 such that both cp and ^p" are uniformly continuous.
Hint: First, for n — 1, map [—1,1] C R onto the line segment
connecting —ei and t\ (the standard unit vectors), then [1,1 + y/2] onto the line
segment connecting e\ to e2, then [1 + \/2,1 -f 2\/2] onto the line segment
between Z2 and e3, etc., and then map (—1, —1 — \/2] onto the line segment
connecting —t\ to — e2, etc. Since the distance of the disjoint line segments
in ?2 we consider is at least -4= and angles between the adjacent line
segments are either \ or ^, we calculate that the inverse map is uniformly
continuous. The map itself is easily seen to be uniformly continuous.
For n > 1, we map ?^ into the product of ^'s, using the preceding map
for coordinates.
10.27 Let Hn be defined as before Proposition 10.36. Show that a set
M C Sx is relatively compact if and only if lim (Hn(x)) = 1 uniformly
n—*oo
for x e M.
Hint: Assume that lim (Hn(x)) = 1 is not uniform on M. Then there are
n—+oo
e > 0, a subsequence {n*} of N and #nfc G M such that Hnk(xnk) < 1 — e.
Since lim (Hn(x)) = 1 for every x E ^x, we get nfc2 > n^ such that
Hnk2(xnkl) > 1 - e/2, then get nkz > nk2 such that Hnk3(xnki) > 1 - e/2
and Fnfc3(xnfc2) > 1 - e/2, etc.
Using the fact that the distance to a subspace is a subadditive function,
we have, for i > j,
\\Xnki — Xnkj || > Mnki\%nkj ~ %nki) > tinki{Xnkj) — -Hnki(Xnki)
> i_|_(i_e) = |.
Therefore, M is not relatively compact. If M is relatively compact, then
M is compact and Hn(x) is a sequence of continuous functions that con-

10. Smoothness and Structure 353
verges to the constant function 1 on M. Hence, the uniformity of the
convergence follows from the classical Dini theorem.
10.28 Let Y be a closed subspace of a Banach space X. Show directly that
X is homeomorphic to Y 0 (X/Y).
Hint: Let ^ be a Bartle-Graves selector; that is, a continuous map from
X/Y into X such that 7t(t/>(#)) = x for every x, where tt is the quotient
map (see, e.g., [BeP2] or [DGZ3]). Define a map <p:X -> Y 0 X/Y for
x G X by
ip{x) — (x — ij){x),x).
For (t/, x) G y 0 X/y, we have </?(y + x) = (y, ?).
10.29 Show that if Y is a separable infinite-dimensional Banach space,
then there exists a C°°-smooth map from ?2 onto Y.
Note that, in general, if X,Y are separable infinite-dimensional
Banach spaces, then there exists a C1-smooth Lipschitz map from X onto
y (Bates).
Hint: Define K — {x = (x{) G Btr\ X{ > 0,J2V^ — ^}- Construct a
bounded linear map T from l\ into X such that |#x C T(K) as follows.
Let {or2} be dense in Bx- Define T(e2) = xl and extend it on t\. Then
T is a bounded linear operator. Given x G Bx, find inductively zn G N
such that HrteJ - ar|| < 2~2 and X] 2-2(n-1)T(et-n) - x
n-l
< 2~2k. Put
y = ? 2-2(-1)ein. Then T(</) - x and Ij, G X, so Ifl^ C T(K).
n = l
Define a map P from ?2 into ^1 by P(ai,a2,...) = (|«i|2, |a2|2, • • •)• Then
K C P(Bt2). The desired operator is Q = T o P.
The next three exercises cover what is called the Gorelik principle, used
in the theory of uniform homeomorphisms. We follow [JLS].
10.30 Let yo be an infinite-codimensional closed subspace of a Banach
space y. Let B be a compact set in Y. Show that for every r > 0 there is
y EY such that \\y\\ < r and dist(J3 + y,Y0) > r/2.
Hint: Consider the quotient map q: Y —> Y/Yq. Then q(B) is a compact set
in Y/Yq. Using a finite ?-net for a small enough ?, we can find y similarly
as we did in Lemma 1.23.
10.31 Let Xo be a finite-codimensional subspace of a Banach space X.
Show that for every r > 0 there is a compact set A C tBx such that if
(p is a continuous map from A into X for which \\<p(x) — x\\ < r/2 for all
x G A, then y>(A) fl X0 / 0.
Hint: Put A = /(fr?x/Xo), where f'\rBx/xQ -> *"?* is the Bartle-
Graves selector (Exercise 10.28). Consider the map x h-> z - 7T(p(f(x)) from

354 10. Smoothness and Structure
\tBxjXq into \tBx/x0- It has a fixed point xo by Brouwer's fixed point
theorem. Then xq — n<p(f(xo)) = xq, and thus ip(f(xo)) G Xo.
10.32 (The Gorelik principle; see [BeLi]) Let U be a uniform homeo-
morphism of a Banach space X onto a Banach space Y. Let Xo be a
finite-codimensional subspace of X, let Yq be an inhnite-codimensional sub-
space of Y, and let 6, d > 0 be such that U(dBx0) C Yo + 6 By. Show that
then d/4 < u(V,26), where V is the inverse map of U and cj(V, •) is the
modulus of uniform continuity of V.
This principle means that a uniform homeomorphism between Banach
spaces cannot carry a large ball of a finite-codimensional subspace into a
small neighborhood of a closed subspace of infinite codimension.
Hint: Put r = d/2 and obtain A from the previous exercise. Consider the
set U(A) and by the second previous exercise get y G Y, \\y\\ < 26, such that
dist(U(A) + y, Y0) > 6. Consider the mapping <? defined from A into X by
(j)(a) = V(Ua + y). Since ||y|| < 26, we have that \\a - </>(a)\\ < u>(V, \\y\\) <
wiy, 26). If lj(V, 26) < r/2 = d/4, by the previous exercise there is ao G A
such that (j>(ao) G A"o- Since in this case also <^(ao) G dBx (ao G A C
r^x = |JBx and \\a0 — ^>(a0)|| < d/4), we have that <f)(a0) G dBx0- Thus
by the assumption, U{4>{aQ)) - U(V(U(a0) + y)) - U(a0) + y G bBYo- This
contradicts dist(U(A) + y, Yo) > 6.
10.33 (Raja [Rajl]) Let X be a Banach space whose norm is LUR. Show
that the norm has a cr-discrete basis consisting of convex sets.
A family of subsets of X is discrete if every point of X has a neighborhood
that intersects at most one member of the family. A family of sets is cr-
discrete if it is a union of a countable number of discrete families.
It is not known whether every Banach space has a ^-discrete basis of its
norm topology consisting of convex sets.
Hint: Fix e > 0 and define by transfinite induction a family of convex sets
{Ba} as follows. Bq = J5x, Ba — f] Bp if a is a limit ordinal, and
Ba+i = Ba \ {x G X] x*a(x) > aa},
where x*a G Sx* and aa G R are such that
diam(?x n {x G X\ x*Q(x) > aa}) < e.
The process ends when B1 C Int(Bx)- Take 6 > 0. Define convex sets
C(a, er, 6) = Ba D {x e X] x*a(x) > aa + 6}.
Then Sx C (J U <?(»,?,<$)• The family {C(a,?,<5); a < 7} is 6-discrete:
<5>0a<7
If a < /?, then for every x G C(a,e1S) and every y G C(/3,e,6) we have
xa(x) >«« + <$ and x*a(y) < aa because y G Ba+i. Thus x*a(x — y) > 6,
and hence \\x — y\\ > S.

10. Smoothness and Structure 355
For n > 3m, define U(a,m,n) = C(a, ^, f) + Int(^Bx). We obtain a
—discrete family {U(a,m, n); a < 7} of open convex sets of diameter less
than —. Prove that
m
{rU(a, m,n)] a < 7, r > 0 rational, m, n G N} U {Int(^jB*); n G N}
is a <7-discrete basis of the norm topology of X: Fix x ? X \ {0}. Take
m > \\x\\/e. For some a and n big enough, ^/||^|| G ?/(a,ra,n). There is
a positive rational r with 0 < r < \\x\\ such that x/r G f/(a,m,n). Thus
? G r[/(a, m, n) and diam(r?7(a, m, n)) < r/m < e.
10.34 Show that every norm on a Hilbert space X is Frechet differentiable
at some points.
Hint: Let || • || be a norm on X and let || • ||x be the Hilbertian norm of X. By
Theorem 10.20, there is / G X* such that inf{||z||?-||z|| + /(?); x e X}-
—6 < 0 is attained at some point x0 G X. Therefore, \\x\\l + f(x) + 6 > \\x\\
for every x G X and ||^o||i + /(#o) + 6 = lko||- Since \\x\\l + /(^o) is
differentiable at xq (f is linear), we have that || • || is differentiable at xo by
Exercise 8.6.
10.35 Find an example of a C2-smooth norm on a Banach space X such
that its dual norm is also C2-smooth but X is not isometric to a Hilbert
space.
Hint: Work in R2.
10.36 Let (X, || • ||) be a Banach space. Show that if || • || is twice Frechet
differentiable at 0, then X is isomorphic to a Hilbert space.
Hint: If ||a:||2 — t(x) +p(x) is the Taylor expansion at 0, note that t(x) — 0
and ||;e||2 —p(x) — o(\\x\\2). Thus there is e > 0 such that p(x) > f |H|2 for
||:e|| > e. Therefore, y/p(x) defines a Hilbertian norm on X.
10.37 Show that all polynomials on cq are weakly sequentially continuous.
Hint: Similar to the proof of Lemma 10.41.
10.38 Let || • || be the canonical norm of ?4. Show directly that || • || is not
twice Gateaux differentiable at the origin whereas || • || is C°°-smooth on
u.
Hint: Direct calculation.
10.39 Assume that a Banach space X has a cr-compact James boundary.
Show that X can be equivalently renormed to have a countable James
boundary.
Hint: Let \JKn be a James boundary of X, where Kn are compact sets in
Sx. For every n G N, fix en > 0 so that A+ ^)(l-?n) > 1+ ^y. Let Fn

356 10. Smoothness and Structure
be an ^-finite net for Kn and F = (J Fn. Define an equivalent norm ||| • |||
on X by |z| - supn{(l + ?) supF»}.
For any x ? X, there is no G N and a neighborhood U(x) of x such that,
for all z E U(x),
sup|(l + ?) sup(z); n > n0\ < sup|(l + ?) sup(z); n < n0>.
From this and the fact that Fn is an e-net for Kn for each n5 there is
/ G |J Fn such that |a:| — f(x). Hence, F is a countable James boundary
n<no
of X in the norm ||| • |||.
10.40 Show that there is an equivalent norm on l\ such that its restriction
to the subspace of all finitely supported vectors in l\ is C^-smooth away
from the origin.
Hint: Let Y be the subspace of l\ formed by all finitely supported vectors.
For n E N, let Bn = {(»»¦) E S^; Xj = 0 for j > n}. Then \JBn is a
James boundary of Y and each Bn is compact. Thus, Y has an
equivalent norm with a countable James boundary and the result follows from
Theorem 10.46, which is also true for normed spaces.
10.41 Show that ?oo/cq does not admit a Lipschitz, Gateaux differentiable
bump function.
Hint: Use Theorem 10.10 and the fact that /?N \ N does not contain any
nontrivial convergent sequence.

11
Weakly Compactly Generated Spaces
Definition 11.1
A Banach space X is called weakly compactly generated (WCG) if there
is a weakly compact set K in X such that X = span(iv).
We say that K generates X if X — span(/<').
If K is ^-compact, then so is K U (-K). Thus, conv{^ U (-K)} is
convex, symmetric, and weakly compact by Theorem 3.58. Therefore, we
may assume in the definition of WCG spaces that the set K is weakly
compact, convex, and symmetric.
Let X, Y be Banach spaces, and let T be a bounded linear operator
from X onto a dense set in Y. If X is weakly compactly generated by a
w-compact set K, then Y is also weakly compactly generated, namely by
the w-compact set T(K). In particular, if Y is a closed subspace of a WCG
space X, then X/Y is a WCG space. A subspace of a WCG space need not
be WCG ([Ros3]; see, e.g., [Fab]).
Examples
(i) Every reflexive Banach space is weakly compactly generated by its unit
ball.
(ii) If X is a separable Banach space and {xn} is a dense set in Sx, then
K — {^#n} U {0} is a compact set that generates X. Therefore, every
separable Banach space is weakly compactly generated.
(iii) If T is a set, {e7} is the family of standard unit vectors in co(r),
and {en} is a sequence in {e7} formed by distinct elements, then en —+
0 pointwise and thus en ^ 0 in cq(T). Hence, by the Eberlein-Smulian

358 11. Weakly Compactly Generated Spaces
theorem, {e7} U {0} is a weakly compact set in cq(T) and cq(T) is thus
weakly compactly generated. Alternatively, the formal identity maps -^(F)
onto a dense set in co(T).
(iv) Since every weakly compact set in ^0 is norm separable (Exercise 3.50),
the space i^ is not WCG. Every weakly compact set in ?i(T) is compact
(Exercise 5.59) and thus separable. Therefore, the space -^i(r) is not WCG
if T is uncountable, since ?i(T) is then nonseparable.
(v) Holder's inequality implies that the identity operator T from L2(a0
into Li(fi), where fi is a finite measure, is a bounded linear operator. From
the theory of the Lebesgue integral, we know that T maps ?2(a0 onto a
dense set in Li(fi). Therefore, ?i(/i) is weakly compactly generated by the
set T{Bl2^). More generally, we obtain that ?i(//) is weakly compactly
generated if // is a-finite.
If [i is not <7-finite, then Li(fi) contains a complemented copy of ?i(T)
for some uncountable T. By (iv), Li(fi) is not WCG.
Definition 11.2
Let X be a Banach space. The density character of X, denoted dens(X);
is the minimal cardinality of a dense set in X.
The weak star density character w*-dens(X*), of the dual Banach space
X* is the minimal cardinality of a w*-dense set in X*.
We have dens(X) < dens(X*) (see the proof of Proposition 2.9). Note
also that w*-dens(X*) < dens(X). Indeed, if {xa} is norm dense in Sx
and fa E Sx* are such that /<*(#<*) = 1, then using Theorem 3.18 we get
conv™ {fa} = Bx* (see also the exercises in Chapter 12).
In general, we do not have equality. For instance, ?\ is u>*-dense in l*^
and thus ^*-dens(^) < Ko < dens(^oo). However, in the case of weakly
compactly generated spaces, we have the following statement.
Theorem 11.3 (Amir, Lindenstrauss [AmLi])
Let X be a weakly compactly generated Banach space. Then dens(X) =
w*-dens(X*).
Proof: Let K be a weakly compact convex subset of X that generates X,
and let D be a set of cardinality tu*-dens(X*) that is uAdense in X*. Define
an operator T:X* -+ C(K) by T(f):k >-+ f(k). Since T{D) separates
points of K, the algebra generated by T(D) and the constant function
contains a dense set N of cardinality w*-dens(X*) = card (T(Z))) that is,
by the Stone-Weierstrass theorem, dense in C(K). For every g E N, choose
kg E K such that g(kg) = sup (#(&))• We claim that S — {kg\ g E N} is
keK
w-dense in K. Indeed, we have sup(#) = sup(^(^5)) for every g E C(K)
K g?N
since otherwise for g E N sufficiently close in C(K) to g we cannot have
that supK(g) = g(kg). Therefore, 5 is weakly dense in K by Urysohn's
theorem, so span(S) is w-dense in X. The set spanE) has a norm dense
subset of cardinality card(S) = w*-dens(X*) (use rational combinations).

11. Weakly Compactly Generated Spaces 359
Since span(S') is weakly and thus norm dense in X, we have dens(X) < w*-
dens(X*). As noted earlier, it is always true that w*-dens(X*) < dens(X).
Alternatively, let a weakly compact set K generate X, and let D be a
set of cardinality tu*-dens(X*) that is w*-dense in X*. The topology r of
pointwise convergence on D coincides with the ^-topology of K because K
is ^-compact, r has a base of cardinality card(D) and thus there is a set S
of cardinality card(jD) that is w-dense in K. Therefore dens(X) < card(D).
?
Corollary 11.4
Let X be a Banach space such that X* is weakly compactly generated. Then
dens(X) = dens(X*).
Proof: By the Goldstine theorem, w*-dens(X**) < dens(X). By
Theorem 11.3, we have dens(X*) = w*-dens(X**). Therefore dens(X*) <
dens(X). As noted above, dens(X) < dens(X*).
?
Recall that u>o is the first infinite ordinal.
Definition 11.5
Let fi > wq be an ordinal A transfinite sequence of bounded linear
projections {Pa}w0<a<fi Z5 called a projectional resolution of identity (PRI) if
Pfi = Ix and for all a,f3<fi we have \\Pa\\ = 1, dens(Pa(X)) < card(a);
PaPp — PpPa = Pmin(o;,/?); and for every x E X the map a *—> Pa(x) is
a continuous map from the ordinal segment [u>o,fj] in its standard topology
into X.
For information on the standard topology of [wo,/i], we refer to [Dug2].
Theorem 11.6 (Amir, Lindenstrauss [AmLi])
Let X be a Banach space weakly compactly generated by a convex symmetric
set K. Let \i be the first ordinal of cardinality dens(X). Then there is a PRI
{Pa}u;o<a<fi such that Pa(K) C K for every a G [wo,A*]-
The proof of Theorem 11.6 will be completed in a series of lemmas, which
are due to Amir and Lindenstrauss.
We will denote the dual norm of (X, || • ||) by || • || again. We will consider
a norm on X* defined by |/| = sup |/(ar)|. Note that | • | < c\\ • || for some
xeK
c > 0 by the boundedness of K. Let B be the closed unit ball of | • |; note
that B = K°. We have the following observation.
Fact 11.7
In the above notation, let T ? B(X*) satisfy \T\ < C for some C > 0. Then
T is the dual operator to some G G B(X).
Proof: Let T*:X** -> X** be the dual operator of T. Then T*E°) C
CB°. By the bipolar theorem, CB° = UK™* in X**, so CB° = CK be-

360 11. Weakly Compactly Generated Spaces
cause K is ^-compact. Therefore T*(K) C CK) and thus T* (span (if)) C
span(iv). Since T* is continuous in || • ||, we have T*(X) = T*(span(if)) C
span(iv) = X. Therefore, T*\x preserves X and it is standard to check
that T=(T*U*.
D
Lemma 11.8
In the above notation, let F be a finite-dimensional subspace of X*, n E N;
and xi,...,xi G X. Then there is a ^-dimensional subspace Z of X*
containing F such that for every closed subspace V of X* containing F
as a subspace of codimension n and every e > 0 there is a linear operator
T:V -> Z satisfying T(f) = / for every f G F, \\T\\ < 1 + e, \T\ < 1 + e,
and \v(x{) — T(v)(x{)\ < e\\v\\ for every i — 1,..., /.
Proof: Let P be a | • [-bounded projection of X* onto F. Since all norms on
a finite-dimensional space are equivalent, there is L > 0 such that \P\ < L
and ||P|| < L. Fix m > 1 such that m > 6A + L)e~l. Let r, s G N. Choose
v\,..., vp G F such that for every v G F, the following two conditions hold:
— If ||t>|| < r, then there is h G {1,... ,p} such that \\v — Vh\\ < ^;
— if |t;| < r, then there is h1 G {1,... ,p} such that |v — tv| < ^.
Choose elements A1,..., \q in St* such that for every A G S*» there is
j G {1,..., q) satisfying ||A — A-7 ||i < ^, where || • [^ is the standard norm
of i™. Consider the following N — 2n + 2pq -f /n real-valued functions of
f=(fu---Jn)?(X*)n:
Wfil \fi\, Ik + E Aj/il, U + E Aj/J,/<(**)>
i=l
* = 1
for 1 < i < n, 1 < h < p, 1 < j < g, 1 < Jb < /.
We can regard them as a mapping $:(X*)n —> ^. Then we find a
sequence {*(/*)}«, f = (/?,. ..,/*)€ (**)"> < G N, which is || • ||TO-dense
in <?((X*)n). Constructing such a sequence for every r, s G N, we obtain
a sequence {/*'r,5}*eN for every r, 5 G N. Let Z be the subspace of X*
spanned by F and {/4-,r,s; tf,r, s E N,i = 1,... ,n}.
Given a closed subspace V D F with dim(V/F) — n, choose w\,..., wn E
(/x - F)G) such that |?A2^-| > ||A||i and ||?A^|| > ||A||i for
every A = (Ai,...,An) E P{. (It is enough to choose w\,...,wn linearly
independent and multiply them by a sufficiently large number.) Choose s
such that \wi\ < s and \\wi\\ < s for all 1 < i < s and then choose r
such that 2 s + 1 < e(r — s). We will fix these values of s and r for the
rest of this proof. Then we can find / = (/i,..., fn) E {/*'r'5} such that
l|$(/i,---,/n)-$(wi,---,Wn)||oo < ?. Define an operator T: V' -» Z by
n n
r(w + J^ A,-u>,-) = v + Yl *ifi

11. Weakly Compactly Generated Spaces 361
for v e F, Xi scalars. Then T: V -> Z and T(v) = v for all v eF.We will
now prove that \T\ < 1 + € and ||T|| < 1 + e. By homogeneity, it suffices
I n I I n |
to show that if ||A||i = 1 then \v + E A*7* < A + e)\v + E ^'^1 and
' 2=1 ' ' 2 = 1
f + E A*7 < A + e) P + E A*^
z = l •' " 2 = 1
because the proof for || • || is the same.
. We will show this for the norm
Assume that \v\ > r. Then
Therefore,
v + E ^'^i > M - E l^f| • |^i| > r - s.
2 = 1 ' 2=1
2 = 1
71 71 74 71
v + ^Af-/i < v + ^Ai^f + HT^i^ + lCA|7"
2=1 2=1 2=1
n
< v + ^Aju;,- +5 + 5 + 1
2 = 1
n n
< v + ^AfixJ + e(r-s) < A + e)\v + ^Al-u;I-
2 = 1
2 = 1
We used the fact that \\wi\ — \fi\\ < ~ < 1 for every i, so EIA*I I/*I <
^ |Af| |w2-| + 1 < 5 + 1. If \v\ < r, let Vh be an —-approximation of v in | • |
and let A; be an ^--approximation of A in ^-norm. We have
v + ^Af/J - v + y^Afti;,-
2=1 2=1
n n
2 = 1
2 = 1
+|DA*-A^|+|DA<-A>--
2 = 1 2 = 1
< JL _|_ _L JL. S±l _L _L_ < A
— m ' m ' ms ' ms — m"
Therefore m + J^ A;/;
I 2 = 1
On the other hand,
<-^ +
— 771 '
V + ? A;Wf
2 = 1
er+y^A*^
2 = 1
- fe^ilSAi""
- 1+L ^ m-
Therefore
v + E Wi < (i + Oh + E A*™*
2=1 I I 2=1

362 11. Weakly Compactly Generated Spaces
We proved \T\ < l + e. It remains to show that \v(xi) — T(v)(xi)\ < e\\v\\.
n
If w = v + ^2 ^iwi and 3 — 1,. - -, /, then
2 = 1
n
W*i)-TW(*:)\ = \52WM*i)-M*i))\ < BMh-
Due to the choice of Wi, we have
1 II
ZTpIT 5>l
\W \\> — —T > XiW.
>
Hence \w(xj) - T(w)(xj)\/\\w\\ < A + L)/m < e.
D
Lemma 11.9
In the above notation, for every finite-dimensional subspace F of X* and
a sequence {xi} C X, there exists a w* -w* -continuous bounded linear
operator T from X* into X* such that \\T\\ = \T\ = 1, T(X*) is a subset of
a w*-separable set in X*, T{v) — v for every v G F, and T*(x{) — Xi for
every i G N.
Proof: Applying Lemma 11.8 to n G N and x\,... ,x\ G X, we find a
^-separable subspace ZU}i of X* such that for every finite-dimensional
subspace V of X* containing F as an n-codimensional subspace there
is a iu*-iu*-continuous bounded linear operator X^: V —>¦ X* satisfying
T^(V) C Zn>/> IT^I < 1 + i, IIT^H < 1 + 1, and Tn^(/) = / for every
/ G F. Put # = spaH^* (|J Znii). Then Tn^( V) C H. Extend the domain of
T^ onto the whole X* by putting T^(X* \ V) = 0 (T^, is no longer linear
on X*). Consider the product space BBx*)Bx*, where 2#x* is
considered in the w*-topology. By Tychonoff's theorem, BBx*)Bx* is a compact
space in its canonical pointwise-t/;*-topology. A net of elements {X^j| }
in BJ3X*)*** is ordered as follows: T^1, < T^2, if Vi C V-*, nx < n2j and
/i</2-
There exists a subnet {T^a l } that converges to some T G BBx*)Bx*.
Extend T homogeneously onto X*. We claim that the extended map T
is linear. Indeed, for arbitrary scalar A and gi,g2 E Bj*, choose 1/ such
that 0!,02 € V and F C V. For arbitrary Tj* /a such that K C 7a, we
have T^lx(\9l(l - \)g2) - ATn^>) - A-\)T?th(g2) = 0. Thus w*-
\im(T^lx(\9l + A - \)g2) - \T?-h{9l) - A - X)T^lxg2) = 0, and T is
linear. Similarly, by taking n\ large, we get \T\ < 1, ||T|| < 1. By Fact 11.7,
T is a dual operator to X*|x. Finally, by Lemma 11.8, we have \v(x{) —
T{v)(Xi)\ = \v(xi-T*(xi))\ < ±\\v\\foTeveryxi,n,veX*,8oT*(xi) = Xi.
D

11. Weakly Compactly Generated Spaces 363
Lemma 11.10
In the above notation, let ft be an infinite cardinal, let F be a closed subspace
of X* with w*-dens(X*) < ft, and let G be a closed subspace of X with
dens(G) < ft. Then there is a w* -w* -continuous bounded linear projection
P in X* such that \P\ = \\P\\ = 1, P(/) = f for every f e F, P*(x) = x
for every x G G, and dens(P(X)) = w*-dens(P*pf*)) < ft.
Proof: By transfmite induction on ft. Assume first ft = ft0. Let {f\} be
w*-dense in F, and let {x\} be dense in G. Then, by Lemma 11.9, there
is a w*-separable subspace F\ of X* and a w*-u;*-continuous bounded
linear operator Tx: X* -* Fx such that |T\| = ||7i|| = 1, Tx{fl) = f{, and
T*(xj) = x\ for every /; hence T({x) — x for every x G G. Let {/?} be a
w*-dense sequence in F\. By Lemma 11.9, there is a ^-separable subspace
F2 of X* and a w*-w*-continuous bounded linear operator T2:X* —» F2
such that ||r2|| = \T2\ = 1, T2(x) = x for every xeG, and T2(f)) = f) for
i, j = 1,2. We proceed by induction on n to get a sequence of w*-separable
subspaces Fn and operators Tn. Their pointwise tu*-limit P (by Tychonoff's
theorem) satisfies P(f) — f on all sequences /• constructed earlier and also
||P|| — \P\ = 1, so P is w*-u>*-continuous. Let F be the tu*-closure of the
span of F and F/s. Then P(f) = f for every / G F; also P(X*) = F and
thus P is a projection.
Now suppose we have proved the lemma for all cardinalities smaller than
ft. Let fi be the first ordinal of cardinality ft. Take a u>*-dense set {fa\ a <
fi} in F and a dense set {xa; a < fi} in G. By the induction hypothesis,
we construct a transfmite sequence of linear projections PQ, a < fi, that
are w*-u>*-continuous and satisfy \Pa\ — \\Pa\\ — 1, w*-dens(Pa(X*)) <
card(a), dens(P*(X)) < card(a), PaPp = Pp, and P*P* = P* if/? < a.
It is routine to check that the pointwise-tu*-limit point of these projections
(in the sense as before) is a projection that satisfies the required properties.
D
Proof of Theorem 11.6: We will use transfmite induction on dens(X).
If dens(X) = fto, then the identity operator satisfies the requirements.
Assume that dens(X) = ft and that the theorem holds for all WCG spaces of
density character less than ft. Let fi be the first ordinal of cardinality ft and
choose a dense set {xa}a<fI in X. By induction, we construct a sequence
Pa of projections as required with xa G Pa+i for each a < //. For the
limit ordinal a, the projection Pa is the pointwise-u>*-limit of projections
{Pp}p<a in the sense as before.
Since P*(f) ^ P*(/) as /? -> a", we have Pp{x) ^ Pa(x) in X.
Therefore |J Pp(X) is dense in Pa(X) for every limit ordinal a. This means that
for each a, dist(Pa(#), Pp(X)) —» 0 as /? —> a. Therefore (Exercise 5.4),
lim (Pp(x)) = Pa(x) in X. Pa(I<) C K follows by duality from \P*\ < 1.
D

364 11. Weakly Compactly Generated Spaces
For an alternative proof of the existence of PRI in reflexive spaces, see
Exercise 11.1. This other method comes from [AMN], [Val2], [Gull], [Gul2],
[Val3], and [OrVa]. A similar method was used also by Plichko in [Pli].
Definition 11.11
Let X be a Banach space. We say that a Markushevich basis {xa\ fa} of X
is weakly compact if {xa} U {0} is a weakly compact set in X.
Theorem 11.12 (Amir, Lindenstrauss [AmLi])
Every weakly compactly generated Banach space admits a weakly compact
Markushevich basis.
PROOF: By transfinite induction on dens(X), we prove that, given a
incompact convex symmetric set K that generates X, there is a Markushevich
basis {xa; fa} of X such that {xa} C K. First, assume that dens(X) = No-
Since K is norm separable, choose {zn} dense in K. By Theorem 6.41, there
is a Markushevich basis {xn; fn} of X such that span{#n} = span{zn}.
Since span{zn} C [JnK, by scaling we can assume that xn E K for every
n
n. Then {xn} U {0} is weakly compact. Indeed, any infinite sequence {xfn}
of distinct elements chosen from {xn} converges to zero in the topology
of pointwise convergence on {/n}, which is a Hausdorff topology. Since
{xn}u{0} C K and K is weakly compact, we have x'n ^ 0. Thus, {^n}U{0}
is w-compact by the Eberlein-Smulian theorem.
Assume that the theorem was proven for every WCG space of density
character less than N. Let {Pa} be projections from Theorem 11.6. Then
(Pa+i — PQ)(X) is weakly compactly generated by (Pa+i — Pa)(K) C 2K.
Using the induction hypothesis, for every a we find a Markushevich basis
{rf]f$}peAaof(P*+i-P*)(X)suchthzt{x°} C (Pa+i-Pa)(K)c2K.
If (Pa+i — P<x)(x) — 0 for every a, then by transfinite induction, Pa(x) = 0
for every a and thus x = 0. Thus, {x%\ Pa(fp)}Q< gG\ ^s a Markushevich
basis of X with {*?}a<M>/,eAB C IK and thus \\x%- lkU$)}a<llMKa «
a weakly compact Markushevich basis with §{*/?}*< /5ga CK'
Corollary 11.13 (Amir, Lindenstrauss [AmLi])
Let X be a weakly compactly generated Banach space. Then there exists a
bounded linear one-to-one operator from X into co(T).
Proof: Let {xa] fa}aer be a Markushevich basis of X. By scaling, we may
assume that ||/a|| < 1 for every a E I\ Define a bounded linear operator
T from X into 4o(r) by T(x) = (fa(x)). Then ||T|| < 1 and T maps
span{xa} into the set of finitely supported vectors in ^(r). In particular,
it maps span{#a} into co(r). Since T is continuous and cq(T) is closed in
^oo(r), we obtain T(X) C co(r.) Since {fa} separates points of X, T must
be one-to-one.
D

11. Weakly Compactly Generated Spaces 365
Definition 11.14
A compact space K is called an Eberlein compact if K is homeomorphic
to a weakly compact set in co(T), in its weak topology.
Example
Every compact metric space is an Eberlein compact. Indeed, if K is a
compact metric space, then K is a w*-compact subset of C(/\)*; moreover,
C(K) is separable (Lemma 3.23). If {/;} is dense in Sc(k)> then the map
T{k) — {jfi(k)} gives a homeomorphism of K onto a compact set in cq.
Theorem 4.50 shows that every Eberlein compact is an angelic space. An
example of a compact space that is not an Eberlein compact is the ordinal
segment [0,u>i]. Indeed, since cji is not a limit of a sequence of ordinals
smaller than u>i, this space is not angelic and thus not Eberlein.
Corollary 11.15 (Amir, Lindenstrauss [AmLi])
Every weakly compact set in a Banach space in its weak topology is an
Eberlein compact.
Proof: Let K be a weakly compact set in a Banach space X. Let Y —
span(A'). Then Y is weakly compactly generated and by Corollary 11.13
there is a bounded linear one-to-one map T from Y into co(F). Since T
is w-iu-continuous from Y to co(r), and since the weak topology on Y
coincides with the weak topology from X, T is a homeomorphism of K
in its weak topology as a subset of X (which is compact), onto a weakly
compact subset of cq(T) in its weak topology.
?
Theorem 11.16 (Amir, Lindenstrauss [AmLi])
Let X be a weakly compactly generated Banach space. Then there is a w*-
w-continuous bounded linear operator from X* into some cq(T).
In particular, Bx* in Us w* -topology is an Eberlein compact.
Proof: Let {xa]fa}a^r be a weakly compact Markushevich basis of X.
Assume that ||/a|| < 1 and define a bounded linear operator T from X*
into Mr) by T(/) = (/(**)).
We claim that T(X*) C co(T). If {xn} is an arbitrary sequence of distinct
elements in {xa}aer, then fa(xn) —» 0 as n —> oo for every a G I\ Since
{^cJaerU {0} is weakly compact and {fa} separates points of X, we have
xn ^ 0 in X. Thus, if / G X* and e > 0, then !/(#<*)| > ? for only a finite
number of ex. Hence T(f) G co(r).
We will show that T is continuous from the w*-topology of X* into the
w-topology of co(r). By Theorem 4.44, it is enough to show that T is w*-
u>-continuous on Bx* • Let fu —> / in Bx* and let ea be a standard unit
vector in ^(r). Then ea(T(fu)) = fv(xa) -+ f(xa) = ca(T(/)). Since
{T(/„)} is bounded in c0(T) and span(ea) = ^(r), we have T(/„) -^ T(f)
in cq(T) and T is w*-u>-continuous.

366 11. Weakly Compactly Generated Spaces
If T(f) = 0 for some / G X*, then f(xa) = 0 for all a G T and / = 0 on
X as span{za} — X. Therefore, T is one-to-one.
?
Theorem 11.17 (Davis, Figiel, Johnson, Pelczyhski [DFJP])
Let X be a WCG space generated by a w-compact set K. Then there is a
reflexive space Y and a bounded linear one-to-one operator T from Y into
X such that K CT(BY).
Proof: We may assume that K is a w-compact, convex, and symmetric
subset of Bx. For n G N, set Un = 2nK + 2~nBx, and let || • ||n be the
Minkowski functional of Un. For x G X, let
H=(ENI»)i-
Let Y = {x G X] ||z| < oo}, and denote BY = {x G Y; |||ar[| < 1}. Let
T be the inclusion map of the normed space Y into X. If x G A', then
||ar||n < 2~n and thus ||z| < 1. Hence K C T{By). We claim that Y is a
reflexive Banach space, T is continuous, and K C T(By), where By is the
unit ball of Y in the norm ||| • |||.
Since Un C 2n+lBx for every n G N, we have || • ||n > 2"(n+1)|| • ||
and thus Y is a normed space. To show that Y is a Banach space, put
Xn = (X, || - ||n) for n G N. Then X n is a Banach space isomorphic to
(OO x
? Xn and define <p: Y - Z by p(y) = (T(y), T(j/),...).
n = l y2
Then <p(Y) = {z ? Z; z = (arn) , j;^ — X\ for all n} is a closed subspace of
Z and (p is a linear isometry of Y onto the Banach space <p(Y). Thus Y is
a Banach space. T can be viewed as a composition of <p and the projection
on the first coordinate in Z, so T is continuous; it is obviously one-to-one.
We claim that T**:Y** -+ X** is one-to-one and that (T**)-\X) = Y.
Indeed, ^**:Y** -> (?X**J takes F G y** to (T**(F),T**(F),...) G
(E^n*J- Since 9? is an isometry into, so is <p** (Exercise 2.39). In
particular, T** is one-to-one. From foO~V00 = Y we Set (T**)^) =
Y.
We will now show that Y is reflexive. Observe that the itAclosure of
T(BY) in X** is T**(?y**)- #-*-? t{By)> then ||z||n < 1 for every n
and thus x e Un for every n. Therefore T(By) C Z7™ in X**. Thus, for
every n we have
T(BYf* C 2nK + 2-nBxW* C 2nK+ 2-nBx**w* = 2nK + 2"n?x~
since /? and #x** are both io*-compact in X**. Using the preceding
remark, we get
T**(BY~) C f]BnK + 2~nBx~) C f)(X + 2~nBx~) = X
n n

11. Weakly Compactly Generated Spaces 367
since X is closed in X**. Thus T**(Y**) C X, so Y** C (T**)^) = Y\
hence Y is reflexive.
D
Corollary 11.18
Every weakly compact set in a Banach space in its weak topology is linearly
homeomorphic to a weakly compact set in a reflexive Banach space in its
weak topology.
In particular, every Eberlein compact is homeomorphic to a weakly compact
set in a reflexive Banach space (in its weak topology).
Proof: Let A' be a weakly compact set in a Banach space Z. Set X —
span(A'). By Theorem 11.17, there is a reflexive Banach space Y and a
bounded linear one-to-one operator T from Y into X such that K C T(By).
Then T~1(Ar) is a weakly closed subset of the w-compact set By, and
thus T~l(K) is weakly compact. Consequently, T~l:K —> T_1(A) is a
homeomorphism in the weak topologies of Z and Y.
?
Corollary 11.19
Let X be a Banach space. X is a WCG space if and only if there is a
reflexive Banach space Y and a bounded linear operator T from Y into X
such that T(Y) is dense in X.
Proof: If Y is reflexive and T is a bounded linear operator from Y onto
a dense subset of X, then X is weakly compactly generated by the weakly
compact set T(By).
The other direction follows from Theorem 11.17.
?
Theorem 11.20 ([AmLi], [Trol], [GTWZ])
Let X be a Banach space.
(i) If X is WCG, then X admits an equivalent norm that is simultaneously
LUR and Gateaux differentiate.
(ii) If X* is WCG, then X admits an equivalent norm \\ • \\, the dual norm
of which is LUR. In particular, || • || is Frechet differentiable.
In the proof, we will use the following lemma.
Lemma 11.21
Define a norm || • || on ti(T) by \\x\\'2 = ||a:||f + ||ar||2, where \\ • ||f denotes
the canonical norms on ?i(T) for i — 1,2, respectively. Then || • || is an
equivalent LUR norm on ?i(T) that is pointwise-lower semicontinuous.
Proof: It is easy to check that || • || is an equivalent norm on ^i(r) that is
pointwise-lower semicontinuous. Let /, /i, /2,... G ^i(r) satisfy
limB||/||2 + 2||/„||2-||/ + /n||2) = 0.

368 11. Weakly Compactly Generated Spaces
Then limB||/||? + 2||/„||? - ||/ + /n||?) = 0 for i = 1,2 (see Chapter 8).
Thus
lim||/„||i = 11/11!. A)
Since || • ||2 is uniformly convex, we get ||/ — /n||2 —> 0, and also
lim(/ - fn)(y) = 0 for each 7 G T. B)
For a subset A of T, we denote by xa the characteristic function of A. For
g G 4(r), define gA = g-XAe *i(r); then ||^r||x = ||^||i + ll/^Hi- If A is
a finite set in T, then from B) we get lim||/^||i = ||/A|| and consequently
A) implies
lim||/^||1 = \\f\%. C)
For n G N and a finite A C T, we also have
||/-/n||l = ||(/-/„)^||1 + ||(/-/„)r^||l
< ||(/-/n)A||l + ||/r\A||l + ||/„rU||l D)
= ||(/ - fn)A\\l + 211/^11! + (H/n^ll! - H/^Hx).
Given ? > 0, choose a finite set A C T such that ||/r^A||i < e. Then, using
B) and C), choose no G N such that for n > no we have ||(/ — /n)A||i < ?
and |||/nU||i - ||/r^||i| < ?• By D), it follows that ||/ - /n||! < 4e for
n > no-
D
Theorem 11.22 (Troyanski)
If a Banach space X is WCG, then it admits an equivalent LUR norm.
Proof: (Godefroy) Let U be a w*-w-continuous one-to-one operator from
X* into c0(r) for some T (Theorem 11.16). Put T = U\ Then T maps c%(T)
onto a norm dense set in X because U is one-to-one and w*-^-continuous.
Moreover, T is i^-w-continuous.
Let || • || be the original norm of X, and let | • | be an equivalent dual
LUR norm on Cq(T) (Lemma 11.21). For n G N and a;El, put
\x\l = mi{\\x-T(g)\\2+^\g\2;gec*0(T)}
OO
and l\x\\2 = J2 2_nkln- This is a norm on x such that III * III < II • II-
n = l
If ||a?|| = 1, then ||a?||n > min{2 /^iTii , |}- This follows by considering
separately cases when |^| > ^ttW and \g\ < 2f|W- Hence> |*|n is an equivalent
norm on X and so is ||| • |||.
Because T(cl(T)) is norm dense in X, it follows that \x\n —¦» 0 for each x G
X. Indeed, given x ? X and e > 0, find g G co(r)* such that ||:c—T(g)\\ < ?.
Then, for n large enough, \x\2n < \\x - T(g)\\2 + ±\g\2 < e2.

11. Weakly Compactly Generated Spaces 369
Furthermore, note that for each x ? X and each n, the infimum in the
definition of |#|2 is attained at some g G cq(T)*. Indeed, given iGl, the
mapping <^:c0(r)* — R defined by </>(g) = \\x - T(g)\\2 + ±\g\2 is w*-
lower semicontinuous and satisfies lim (<f>x(g)) = oo. Hence <j>x attains its
minimum on co(r)*.
We will now show that ||| • ||| is LUR. Assume that x: xi, X2,... G X* are
such that 21sy I2 + 2|a:|||2 - \\x + Xjf -> 0. Then, for every n, 2|x|2 +
kiln ~ \x + xj\n —> 0 as j -» oo. Find g,gj G Cq(T) such that |z|2 —
Ik - T(9)\\2 + ^W \*i\l = \\*i ~ Tfo)ll2 + ?lfcf- Then
2|*|'+2|ar,-|S-k + ari|2
> 2\\x-T(g)\f + ±\g\2 + 2\\xj-T(gj)\f + l\gjf
-Wx + Xj-Tig + gjW-ftg + gtf
> (\\x - T(g)\\ - H*,- - T(gj)\\)' + IB|ff|2 + 2|ffi |2 - |<7 + Si |2).
This implies that ||ay-r(#)H - ll*-r(ff)|| and 2|<7|2+2|5i|2-|<H-Srj|2 — 0.
Because | • | is LUR, we have \g — gj \ —> 0. Thus, for all n, lim sup \\x — Xj || <
limsupdl* - T(ff)|| + \\T(g - 9j)\\ + \\Xj - T(9j)\\) = 2||z - T(g)\\ < 2\x\n.
Since \x\n —» 0, we get \\x — Xj\\ —> 0 and thus ||| • ||| is LUR.
?
Proof of Theorem 11.20: (i): Let X be WCG, and let T be a bounded
linear one-to-one operator from X* into Co(F) that is u>*-tu-continuous
(Theorem 11.16).
Since co(T) is WCG for every T, it admits an equivalent LUR norm ||| • |||
by Theorem 11.22. Let || • || denote the original dual norm of X* and define
anorm|.|onX'byJ/|2 = ||/||2 + |T(x)|2.
Then | • | is an equivalent w*-lower semicontinuous norm on X* that is
strictly convex. Indeed, if for some f,g G X* we have 2|/2+2|y|2 —1/+^|2 =
0, then by convexity 2|[r(^)|||2 + 2|||T(y)|i2- |||T(/)-fT(<7)|||2 = 0. Since ||| • |||
is strictly convex on co(r), we get T(f) = T(g) and thus f — g because T
is one-to-one.
In order to obtain the w*-lower semicontinuity in (ii), it suffices to note
that the norms are w*-lower semicontinuous and the infima in question are
attained.
The combination of norms needed to get the rest of the proof follows the
proof of Theorem 8.20.
?
Theorem 11.23
Let X be a Banach space. The following are equivalent:
(i) X admits a shrinking Markushevich basis.
(ii) X is WCG and admits an equivalent Frechet differentiable norm.

370 11. Weakly Compactly Generated Spaces
Proof: (Sketch) (i) => (ii): Let {xa\ fa}aer be a shrinking Markushevich
basis of X, {xa} bounded. Let {yn} be a sequence of distinct elements of
{xa}. Then fa(yn) —+ 0 for every a and thus yn A 0. Hence {#a} U {0} is
a w-compact set in X by the Eberlein-Smulian theorem. Therefore, X is
weakly compactly generated.
Assume that {xa\ fa} is a shrinking Markushevich basis of X such that
{fa} is bounded. Define a bounded linear operator T: X —» co(T) by T(#) =
(/a(a?)). Let ea, a G I\ denote the standard unit vector in ^i(F). Then
T*(ea) = fa for each a G T and thus T* maps ^i(r) onto a norm dense
set in X*. By the proof of Theorem 11.22, X* admits an equivalent dual
LUR norm and thus X admits an equivalent Frechet differentiate norm
by Fact 8.18.
(ii) => (i): Assume for simplicity that dens(X) = Ni and that the dual
norm || • ||* is LUR. Let {Pa}a<u1 be a PRI for the norm || • ||
(Theorem 11.6). Then, by the proof of Proposition 8.34, {P*} is a PRI for
(X*, || • ||*). Because all Pa(X) are separable and thus P*(X*) is
separable, we get that each (Pa+i — Pa)(X) admits a shrinking Markushevich
basis by Exercise 6.37. By a standard "gluing together" argument (see
Theorem 11.12), we obtain a shrinking Markushevich basis of X.
?
Corollary 11.24 ([JoZl])
Let X be a Banach space. If X admits a shrinking Markushevich basis, then
so does every closed subspace of X.
Proof: Every closed subspace of a weakly compactly generated Banach
space admits a projectional resolution of identity (see, e.g., [DGZ3]). Then
we can proceed as in the proof of (ii) => (i) in the preceding theorem.
?
Note that there is a non-shrinking Schauder basis {xn} of a separable
Banach space such that {xn} U {0} is weakly compact ([PeSz]).
Weakly Compact Operators
Definition 11.25
Let X,Y be Banach spaces, T E B(X,Y). T is called weakly compact if
T(Bx) is a weakly compact set in Y.
Let X,Y be Banach spaces. If X is reflexive, then every bounded linear
operator T from X into Y is weakly compact. Indeed, since T is w-w-
continuous (Exercise 3.19) and Bx is w-compact, T(Bx) is ^-compact in
y. On the other hand, the identity operator Ix on a nonreflexive Banach
space X is not weakly compact because Bx is not ^-compact.

11. Weakly Compactly Generated Spaces 371
Lemma 11.26 (Grothendieck)
Let X be a Banach space, A C X. If for every e > 0 there is a w-compact
set Ae C X such that A C A? + eBx, then A is relatively w-compact.
Proof: Since A is bounded, it is enough to show that the w*-closure of
A in X** is actually in X. Since A? and eBx** are both w*-compact,
A? -f sBx™ — A? -f eBx**- Moreover, X is closed in X**, so J" C
0 (A? + eBx** )cf](ln eBx**) = X.
a
Note that the space of all weakly compact operators from a Banach space
X into a Banach space Y is a closed subspace of the Banach space B(X, Y).
Indeed, assume that Tn are weakly compact and Tn —> T in B(X, Y).
Given e > 0, there is n0 such that T(BX) C Tno(?x) + effy. Therefore
T(Bx) C TnoEx) + ?#y and the latter set is closed, since Tno(Bx) is
w-compact and s?y is closed. Thus T(BX) C TnoEx) + ?#y, and TEx)
is w-compact.
Note also that a composition of a weakly compact operator with a
bounded linear operator in any order is weakly compact. In particular,
consider a bounded linear operator T ? #(X, Y). If there is a reflexive
space Z and operators R ? B(X, Z), S ? B(Z,Y) such that T — SR} then
T is weakly compact.
The following theorem shows that factorization through reflexive spaces
characterizes weakly compact operators.
Theorem 11.27 (Davis, Figiel, Johnson, Pelczyriski [DFJP])
Let X, Y be Banach spaces. IfT is a weakly compact operator from X into
Y, then there is a reflexive Banach space Z and bounded linear operators
R:X -* Z, S:Z ^Y such that T = SR.
Proof: Assume without loss of generality that T(X) — Y; otherwise
replace Y with T(X). Then T(Bx) is a weakly compact, convex, and
symmetric set that generates Y. From the construction of a reflexive space Z
in Theorem 11.17, it follows that T maps X into Z because T(Bx) C B%.
Define S(x) — T(x) ? Z and for R take the operator denoted by T in the
proof of Theorem 11.17.
?
As a corollary, we have the following result.
Theorem 11.28 (Gantmacher)
Let X, Y be Banach spaces, T ? B(X,Y). T is weakly compact if and only
ifT* is weakly compact.
Proof: If T is weakly compact, by Theorem 11.27, there is a reflexive
Banach space Z and R ? B(X,Z), S ? B(Z,Y) such that T = SR. Then
T* = R*S* and R*:Z* —» X* is weakly compact because Z* is reflexive.
Thus, T* is weakly compact.

372 11. Weakly Compactly Generated Spaces
If T* is weakly compact, then T**: X** —> Y** is weakly compact by the
first part of the proof. Since T**|x = T, we have T(BX) = T**(BX) C
T**(BX**): Since T**(Bx**f is ^-compact in Y**, we have T(BX)W C
T**(BX**) and T(Bx) is ^-compact in Y**, and thus ^-compact in Y.
D
Definition 11.29
Let X,Y be Banach spaces, T ? Z?(X, Y). T is called absolutely summing
tfYl ||2^C?*)||v < °° whenever ^2,xi is unconditionally convergent in X.
In other words, T carries every unconditionally convergent series in X
to an absolutely convergent series in Y.
Examples
(i) The identity operator J on ?2 is not absolutely summing. Indeed, let {e,-}
be the canonical basis of ?2. Then J2 Tei is unconditionally convergent, but
EllHl2 = ?H°°-
(ii) The formal identity operator / from C[0,1] into Li[0,1] is absolutely
summing. Recall that if xi,...,xn are vectors in a Banach space ? X,
then sup< J2 eixi\'> ei = ±1 f = sup< ]P |ar*(a;z-)|; a:* ? Sx* \ (note after
Theorem 6.38). Using this, for /1,..., fn G C[0,1] we get
Eii/iiii = e/1i/<(oia=/1Eimoi*< »p Eimoi
fcl ,-=i-A> -/O ,.=1 *6[0,l]i=1
n n
< sup Vl^/OI-supllV^/JI ;ei = ±lj.
If ^2 fi ls unconditionally convergent in C[0,1], then
n
supjsup ^e:t-/i ; a = ±l| = M
2 = 1
n
is finite (Exercise 1.38). Therefore ]T) ||/z-||i < M for all n, which means
t=i
that I is absolutely summing.
(iii) Every bounded linear operator from ?\ into ?2 is absolutely summing
(Grothendieck; see, e.g., [LiT2]).
Theorem 11.30
Let X,Y be Banach spaces, T ? #(X, Y). IfT is absolutely summing, then
T is weakly compact.
If, moreover, X is reflexive, then T is a compact operator.
Before proving Theorem 11.30, we state one corollary. It shows that
finite-dimensional spaces are characterized by Riemann's result that all
unconditionally convergent series converge absolutely.

11. Weakly Compactly Generated Spaces 373
Theorem 11.31 (Dvoretzky, Rogers; see, e.g., [LiT2])
In every infinite-dimensional Banach space, there is a series that is
unconditionally convergent but not absolutely convergent.
Proof: If in some Banach space X every unconditionally convergent series
is absolutely convergent, then the identity map Ix is an absolutely summing
operator. By Theorem 11.30, Ix is weakly compact, and thus Bx is weakly
compact and X is reflexive. By the second part of Theorem 11.30, Ix is
a compact operator, which means that Bx is compact and X must be
finite-dimensional.
?
To prove Theorem 11.30, we will need the following lemmas.
Lemma 11.32
Let X,Y be Banach spaces, T G B(X>Y). IfT is absolutely summing, then
there is a constant K > 0 such that for all n G N and vectors {?«}._, in
X we have
J2\\nxi)\w<K sup {x>*(*oi}-
2 = 1 ^ A 2 = 1
Proof: Let Z be the space of all sequences {xi}^ in X for which the series
Y^xi is unconditionally convergent, with the norm sup < ^ |?*(#j)| >.
x*6Sx* S' = l J
Similarly as in the proof of Proposition 1.11, we find that Z is a Banach
space. If {x{} G Z, then ]P |\T(xi)\\y < oo. The Baire category argument
gives that there are K' G N and c > 0 such that ^ ||T(x2)||y < K'
whenever {xi} G Z is such that sup < Yl \x*(xi)\ > < c. A standard
**€SX* S'=l ^
homogeneity argument then gives the existence of K > 0 such that
oo oo
^||T(^-)l|y<^ sup {5>*(*)l}-
Now, given a?i,..., xn, we can choose xn+i = xn+i = • • • = 0 and use the
preceding estimate.
D
A bounded linear operator is absolutely summing if and only if it satisfies
the conclusion of Lemma 11.32 for some K > 0. The smallest possible
K > 0 is called the absolutely summing norm of T.
Lemma 11.33 (Pietsch; see, e.g., [LiT2])
Let X,Y be Banach spaces, T G B(X,Y). IfT is absolutely summing, then
there is a regular probability measure \x on (Bx*, w*) and a constant K > 0
such that for every x we have
\\T(x)\\y<k[ \x*(x)\d(i(x*).
v'Bv.

374 11. Weakly Compactly Generated Spaces
PROOF: Consider (Bx*,™*)- We treat X as a subspace of C(Bx*)] in
particular, for x G X we have the function \x\:x* »—> |x(x'*)| for x* G
Bx* ¦ We may assume that, for every n G N and x\,..., xn G X, we have
? ||r(xt-)||y < sup {J \x*(xi)\\. Define F^ C C(?**) by Fx =
t = l a:*G5x* S"=l J
{/ <= C(BjtO; suP (/(**)) < 1} and ^2 = conv{|*|; x G X, \\T(x)\\Y =
l}. Then F\ and F2 are convex subsets of C(Bx*) and F\ is open.
We claim that F\ fl F2 = 0. Indeed, take / G ^2- By the definition of
JF2, there are i,'Gl,i=lr.., n, with ||T(x2)||y = 1 and A; G [0,1] with
J2 A,- = 1 such that /(**) = ? A,-|ar*(a:,-)| for ?* G ?**. Then
*=1 i=l
n n
sup (f(x*)) - sup {VA2|x*(^)|} = sup {y^k*(A2a:2)|}
x*ebx* x*eBx* ^ J x*eBx* ^fr[ >
> ^imA^)l|y-EA,||T(^)ll = l-
z=l i=l
Therefore / g Fi.
By the separation theorem (and using 0 G Fi), there is $ G C(/i)* such
that $(/) < 1 for every / G Fi and <?(/) > 1 for every / G F2. We claim
that if g G C(Bx*), g > 0, then $(G) > 0. Assume the contrary. Then, for
/? negative but large in absolute value, we would have <?(/?#) > 1, but also
fig < 0. Hence Cg G Fi, so $(/?#) < 1, a contradiction.
Therefore, 3> is a nonnegative functional, and by the Riesz representation
theorem, there is a positive regular measure fl on Bx* that represents <$.
Since Fi contains the open unit ball of C(Bx*), we have ||$|| < 1 and
thus fl(Bx*) = $(xbx*) — $(lim((l — ^)xbx*)) < 1- Moreover, for every
/ G F2 we have fB ^ f dfl = $(/) > 1, and thus the measure /i = ft/fl(Bx*)
is a probability measure on Bx* such that /i(/) > 1 for every / G F2. Thus,
if x G X is such that ||T(a?)|| = 1, then |x| G F2 and hence
\\T(x)\\Y = 1 < / \x\dpL= f \x*(x)\ dfi(x*).
J Bx* J B x*
A standard homogeneity argument concludes the proof.
?
Proof of Theorem 11.30: Let x G X. Consider x as an element of
L>2{Bx*, a0, where /i is the Pietsch measure from Lemma 11.33. By Holder's
inequality, we then have
l|r(*)llr < Kl \x(x*)\dti(x*)
Jbx.

11. Weakly Compactly Generated Spaces 375
< K([ \x(x*)\2dti(x*)Y=K\\x\\L2lBxmttly
Hence we may consider T a bounded linear operator (also w-ii'-continuous)
from the reflexive space span'IH2(X) C L2(Bx* , aO into Y. Thus, T(Bx) is
a weakly compact subset of Y.
Now assume that X is a reflexive Banach space. Given a sequence {xn}
in Bx, by ^-compactness we may assume that xn —> x E Bx in X. Then
?n and x considered as continuous functions on (Bx*, w*) are all bounded
in absolute value by 1 and xn(x*) —> x(x*) for every x* E #x* • By the
Lebesgue dominated convergence theorem used for the Pietsch measure /i
from Lemma 11.33, we have that xn —»¦ x in L\(Bx*, aO- By the Pietsch
inequality, ||T(xn—#)||y < J^ |xn —x| d/i —» 0. Therefore, T is a compact
operator.
n
Consider an operator T E 5(X, y). Let i be the natural embedding of
X into the space C(Bx* ,w*)- Denote Xqq — i(X). Let /i be a probability
measure on (Bx*,w*)- Consider a norm-one inclusion i\ of C(Bx*, w*)
into Li(Bx*,fJ>) and denote Xi = 2i(Xoo) • Using Lemma 11.33, one can
show that the operator T is absolutely summing if and only if there exists
an operator S:Xi —*Y such that T = Si\i.
Thus, absolutely summing operators are characterized by this
factorization. This is why Lemma 11.33 is often called the Pietsch factorization
lemma. For more information on the theory of absolutely summing
operators and the related field of the local theory of Banach spaces, we refer to
[Pis2] and [T-J].
Definition 11.34
Let X be a Banach space. We say that X has the Dunford-Pettis property
if x*n(xn) —» 0 whenever xn E X and #* E X*, n E N, satisfy xn —» 0 in
X and z* ^ 0 in X*.
An example of a Banach space with the Dunford-Pettis property is l\.
Indeed, if xn —> 0 in l\, then xn —* 0 by the Schur property of l\. If,
moreover, z* E l\ are such that z* -^ 0 in t\, then sup ||#* || < C < oo for
some C > 0 and thus |ar*(arn)| < ||#* || ||arn|| < C||#n|| ~+ 0.
It is easy to see that X has the Dunford-Pettis property if X* does.
Thus Co has the Dunford-Pettis property.
Proposition 11.35
Let X, y be Banach spaces, T E B(X)Y). Lf X has the Dunford-Pettis
property and T is weakly compact, then T is completely continuous.
Proof: By contradiction: Assume that for some 6 > 0 and xn —> 0 we have
||T(xn)|| > 6 for all n. Let y*n E Sy* be such that y^(T(xn)) = \\T(xn)\\ for
all n. Since T* is weakly compact, by passing to a subsequence if necessary,

376 11. Weakly Compactly Generated Spaces
we may assume that for some x* 6l*, T*(y*) ^> x* E X* in X*. Since X
has the Dunford-Pettis property, we have
0 = lim(T*(»;) - **)(*„) = lim(j?(r(*n)) - **(*„)) = lim||T(a:n)||
as lim(a?*(a?n)) = 0. This contradicts ||T7(xn)|| > 6 > 0 for all n.
n
One consequence is that an infinite-dimensional reflexive Banach space
cannot have the Dunford-Pettis property. Note that the identity map Ix
of a reflexive Banach space X is weakly compact, so if X had the Dunford-
Pettis property, by Proposition 11.35 it would be completely continuous.
Every completely continuous operator from a reflexive Banach space into a
Banach space is compact (Eberlein-Smulian theorem), and thus Ix{Bx) —
Bx would be compact, a contradiction.
Theorem 11.36 (see, e.g., [DiUh])
Let K be a compact space. Then C(K) has the Dunford-Pettis property.
Proof: Assume that fn € C(K), fn^O'm C(K) and Fn e C(K)*,
Fn^O'm C(K)\ Set A = sup ||/n|| < oo and B = sup ||Fn|| < oo.
We identify Fn with regular measures on K by the Riesz representation
theorem, and by the Hahn decomposition theorem we assume without loss
of generality that Fn are nonnegative measures. Since Fn —* 0, there is (see,
e.g., [DuSc]) a positive measure fi on K such that the sequence {Fn} of
measures is equiabsolutely continuous with respect to /i; that is, for every
e > 0 there is 6e > 0 such that \Fn(U)\ < e for every n whenever U C K
satisfies n(U) < S?.
We will show that Fn(fn) —> 0. Given e > 0, get 6? > 0 from the
equiabsolute continuity of Fn. By Egorov's theorem, there is U C K such
that //([/) < 6? and lim (/n) — 0 uniformly on K \U. Thus, there is no
n—+oo
such that for n > no and every t ? K\U we have |/n(*)| < s. Then, for
n > no, we have
\Fn(fn)\=\[ fndFn\< [ \fn\dFn+ f \fn\ dFn < SA + SB.
{Jk ' Ju Jk\u
n
Proposition 11.37
Let X be a Banach space with the Dunford-Pettis property. If Y is a
complemented subspace of X} then Y has the Dunford-Pettis property.
Proof: Let P be a bounded linear projection of X onto Y. Let yn ?Y,
yn ^ 0 in Y and j? e Y*, y*n ^ 0 in Y*. Then yn ^ 0 in X and P*(^) ^ 0
in X* because P* is w-w-continuous. Since X has the Dunford-Pettis
property, P*{y*n){yn) - 0. Hence y*n(yn) = y*n(P(yn)) = P*(y*„)(yn) - 0.
D

11. Weakly Compactly Generated Spaces 377
Corollary 11.38
If Y is a reflexive complemented subspace of C[0,1], then Y is finite-
dimensional.
Proof: By the preceding proposition, Y has the Dunford-Pettis property.
By the remark following Proposition 11.35, every reflexive Banach space
with the Dunford-Pettis property is finite-dimensional.
?
Quasicomplements
Definition 11.39
Let Y be a closed subspace of a Banach space X. A closed subspace Z of X
is called a quasicomplement ofY in X ifYHZ = {0} and Y + Z is dense
in X.
If such a quasicomplement exists, Y is called quasicomplemented in X.
Theorem 11.40 (Murray [Mur], Mackey [Mac])
Let X be a separable Banach space. Then every closed subspace of X is
quasicomplemented in X.
Proof: (Gurarii-Kadec) Let Y be a closed subspace of X. Let {x2;/2} be
a Markushevich basis of Y and {{?;}U{2:;}; {^}U{</>/}} its extension to a
Markushevich basis of X (Theorem 6.42). Put Z — span{z2}. Let w G YC)Z.
Since w G span{z2}, we get ipi(w) = 0 for every i. Since w G span{xz}, we
also get <pi(w) — 0 for every i. Then w = 0 because {ipi}U {<?>;} separates
points in X. Moreover, Y + Z D span{{#2-} U {?{}} = X.
n
In fact, a more general result is true.
Theorem 11.41 (Lindenstrauss [Lin7], [Lin5])
Let X be a weakly compactly generated Banach space. Then every closed
subspace of X is quasicomplemented in X.
Theorem 11.42 (Rosenthal [Rosl])
The space c$ is quasicomplemented in ?^ •
Proof: (Sketch) Since /?N \ N is perfect, there is a continuous map <p of
/?N \ N onto [0,1] ([Lac2]). Therefore, C[0,1] is isomorphic to a subspace
of C(/?N \ N)* = Cq C i^Q. Because Za[0,1] is isomorphic to a subspace of
C[0,1]*, it is isomorphic to a quotient of C(/?N \ N)* and thus by
Rosenthal's result in [Rosl], Li[0,1] is isomorphic to a subspace of C(/?N \ N)*.
Since ?2 is isomorphic to a subspace of Li[0,l] by Theorem 6.28, ?2 is
isomorphic to a subspace of Cq C ?%q ; call this subspace H\.
For n e N, define 6n G t^ by 6n(f) = /(n). Let H - span{^ + fin] n G
N}, where {/in} is a Schauder basis of Hi equivalent to the canonical basis

378 11. Weakly Compactly Generated Spaces
of ^2- It follows that there are constants K\, K<i > 0 such that
m i_ m m \_
n = l n = l n — l
for all ra E N. Thus H is isomorphic to ?2} and therefore # is w*-closed
in I^q (Lemma 6.44). We will show that H±_ is a quasicomplement of c$ in
?oq. By duality, it suffices to show that H C\c$ — {0} and H± fl Co — {0}.
Take f ?cQnHL. Then (/in + ^)(/) = 0 for all n. Because /in t c0 , we
get (^)(/) = ?/(n) = 0 for all n and thus / = 0.
Now take y E H C\ Cq. There are real numbers a; such that J2 \ai\2 < °°
oo oo
and y = Yl an(^ + ^)^ Since fin E cfr for all n, we have y- J2 ctnfjin G Co"
n=l n=l
oo
and thus J2 <*n( „ ) E Cq". For n E N, define a function en on N by
n = l
en(m) = <5nm (Kronecker delta). Then Y^ak(k(en)) = lt- Hence an = 0
for all n, so y = 0.
?
Theorem 11.43 (Lindenstrauss [Lin6])
IfT is uncountable, then co(T) is not quasi complemented in ^(r).
Proof: By contradiction, assume that Z is a quasicomplement of co(T)
in ^oo(r). Let q be the quotient map of ^oo(F) onto ^(r)/^, and
consider T = q\ frv T(c0(T)) is dense in lJ[T)/Z and thus ?oo(T)/Z is
weakly compactly generated. Hence B^^yzy is w*-sequentially compact
(Theorem 11.16 and Theorem 4.51 in co(r)). We claim that q* is weakly
compact.
Let yn — q*(xn), xn E By^ryz)*, and let xnk be a w*-convergent
subsequence of {xn}. Then {ynk} is a u>*-convergent sequence in ^(r). By
the Grothendieck property of ^oo(r), we have that {ynk} is iu-convergent.
Hence q* is a weakly compact operator and so is q] the same is then true
forT.
Therefore, T*: (t^iY)/Z)* —> ?i(T) is a weakly compact operator and
thus a compact operator because ?\(T) has the Schur property. Hence,
T*((?oo(T)/Z)*) is a separable subset of ^i(r). Since T is one-to-one (from
the definition of the quasicomplement), T* maps ?oo(T)/Z onto a u>*-dense
set in Cq(T). This means that Cq(T) is w*-separable, a contradiction (see
Proposition 11.3).
?
Theorem 11.42 was later generalized as follows.
Theorem 11.44 (Johnson [Joh2])
Let Y be a closed subspace of a Banach space X. If Y* is w*-separable

11. Weakly Compactly Generated Spaces 379
and X/Y has a separable and infinite-dimensional quotient, then Y is
quasicomplemented in X.
We remark that the problem of quasicomplementation is closely related
to the problem of the existence of nontrivial separable quotients. Indeed, as
shown by Rosenthal in [Rosl], X admits an infinite-dimensional separable
quotient if and only if X has an infinite-dimensional separable
quasicomplemented subspace. We refer the reader to [Muj] for a survey in this
area.
Exercises
11.1 Let Y be a separable subspace of a reflexive Banach space X. Follow
the hint to show the existence of a norm-one projection on X such that
P(X) is separable and contains Y.
Hint: Let JVi be a separable 1-norming subspace for Y, and let M\ be a
separable 1-norming subspace for JVi, M\ D Y. Let N2 D Ni be a separable
1-norming subspace for M\, M2 D M\ a separable 1-norming subspace for
N2, etc. Put M - \jMi and N = \jNi. If x G M and y G Nj_, then
\\x+ y\\ > \\x\\ (see tne proof of Theorem 6.14). Hence M H N± = {0} and
M + N± is closed in X. If / G ML n TV, then / = 0 since M norms N.
Hence M + A/j_ — X and the map a? -f y —> a? for x ? M and y G A^j_ is the
desired projection.
11.2 Let C be a w-compact set in a Banach space X. Show that if X* is
u>*-separable, then C in its ^-topology is metrizable.
In particular, C is separable. Thus we get an alternative proof that a
WCG space X is separable if X* is tu*-separable.
Hint: Proof of Propositions 3.26 and 3.28.
11.3 (Johnson-Lindenstrauss) Let Y be a closed subspace of a Banach
space X such that X/Y is separable. Show that X is WCG if and only if
Y is WCG.
There are WCG Banach spaces with closed subspaces that are not WCG
([Ros3]). It is not known whether X is WCG if X** is WCG.
Hint: Assume that Y is WCG. Let {xn} be dense in X/Y. Pick xn G xn
with ||xn|| < 2. Let K be a ^-compact set generating Y. Then {-zn} UA'
is a w-compact set generating X.
Assume that X is generated by a weakly compact Markushevich basis
{%a}aer- Then To = {a; xa ^ 0} is countable. Indeed, otherwise, since
X/Y is separable, there would be nonzero condensation points of {xa; a G
To} in the tu-topology, which contradicts the fact that for every sequence of
distinct points {xn} in {xa} we have xn —> 0. Let Z — span{#a; a (fc To}.

380 11. Weakly Compactly Generated Spaces
Then Z is a WCG closed subspace of Y. Y/Z is separable since X/Z is
separable. By the first part, Y is WCG.
11.4 Let y be a closed subspace of a Banach space X. Show that if Y is
reflexive and X/Y is WCG, then X is also WCG.
Hint: Let {xa} be a weakly compact Markushevich basis of X/Y. Choose
X & KZ. X(X)
\\xa\\ < 1. We claim that {xa} U {0} U By is weakly compact.
Indeed, if yp G {xa} and yp —> y ? ^**> then 2/0 "^ 0 m -X/Y. If <Z denotes
the quotient map X -» X/Y, then g**(y,0 ^ g**(y) in X**/Y™\ Since
—w*
yp —+ 0, we have q**(y) = 0. Therefore y EY = Y because Y is reflexive.
Thus y E -By, and the rest is standard.
11.5 Prove that for every separable Banach space X there is a compact
K C Sx such that X — span(iif). What about such a weak compact set
for WCG spaces?
Hint: Let X — \jFn, where Fi C ft C ... are finite-dimensional. By
induction, find a convergent sequence {xk} in Sx such that Fn C span{#fc}
for every n.
The answer for the second part is negative if the space is nonseparable,
it is enough to take any space on the unit sphere of which the norm and
weak topologies (and hence compacts) coincide.
11.6 Show that if X* is WCG, then X is an Asplund space.
Hint: If Y is a separable subspace of X, then Y* is a quotient of WCG and
thus WCG. Then Y* is separable since Y** is w*-separable.
11.7 Show that i^ is not a subspace of any WCG Banach space.
Hint: Assume that i^ C X and X is weakly compactly generated by a
weak compact K. Let P be a projection of X onto l^ (Proposition 5.13).
Then l^ is weakly compactly generated by P(K), a contradiction.
11.8 Does there exist a bounded linear operator from cq(c) onto a dense
subset of -?oo?
Hint: No, i^ is not WCG, whereas Cq(c) is.
11.9 Is there a bounded linear operator from ^ onto a dense set in ^(c)?
Is there a bounded linear operator from l^ onto a dense set in cq!
Hint: Yes, ^(c) is a quotient of l^ (follows by reflexivity from ?2@) C ^,
[Ros2]). Then use the formal identity map from ^(c) into co(c).
11.10 It is known that ?00 (r) does not admit an equivalent strictly convex
norm ([DGZ3]). Use this to show that if T is uncountable, then there is no
bounded linear one-to-one operator from i^(T) into cq(T).
Hint: Theorems 8.13 and 11.22.

11. Weakly Compactly Generated Spaces 381
11.11 We proved that, as a WCG space, c0(r) admits an equivalent LUR
norm. Consider the following "Day's norm": For x = (z7) ? co(r), we define
11 a: 11 = sup< f J2 -^M k where the supremum is taken over all n ? N and
all ordered rc-tuples G1,..., jn) of distinct elements of T.
Show that this norm is strictly convex. In fact, it is also LUR ([DGZ3]).
Hint: If m > n and \a\ < |6|, then ^ + ^ < ? + ?. Thus ||z|| =
/ OO r2 1
( ?^ -^ J , where jj are distinct and such that |a?7l | > \xl2 \ > .... Calling
such a sequence {7^} an appropriate sequence for x, we have that if \\x +
y\\ = ||#|| + ||2/||, ||#|| = II2/H = 1, and {jj} is an appropriate sequence for
x + y, then
2 = „I+„li=(f;^)*
i=i i=i j=i j=i
where {p^} is an appropriate sequence for x\ a contradiction unless {7^}
is an appropriate sequence for x. Similarly, we argue for y. Thus jj is an
appropriate sequence for both x and y. By the parallelogram equality, we
then get x — y.
11.12 Let X be a WCG Banach space X. Show that if cq is a subspace of
X, then Co is complemented in X.
Hint: There is a complemented separable Z C X such that cq C Z. Use the
Sobczyk theorem in Z.
11.13 Let {X^^er be WCG Banach spaces. Then (Z,xh)p is WCG if
p ? (l,oo) and (?X„)Co(r) is WCG. Also, (?***)*!(r) is WCG if T is
countable. Prove these statements.
Hint: Let p ? (l,oo) and K^ be w-compact, symmetric, convex, and
generating Xp, then the set K = {(x^) ? (%2xp)p> xv ? ^>II(xa*)II ^ 1)
is ^-compact and generates (YlX^)p- The space {Y^X^ is mapped onto
a dense set in (Y,x^)c0(r) by the formal identity operator. Finally, if Xn
are weakly compactly generated by lu-compact symmetric convex sets Kn,
then K = |(xn) ? i^Xn)x\ xn ? ifn, ||(arn)|| < ^ j is a weakly compact
set which generates (]PXn)i.
11.14 Show that if X is a separable Banach space, then we have card(X) —
card(X*).

382 11. Weakly Compactly Generated Spaces
Hint: Continuous functions are determined on a countable set. For nonsep-
arable X, even density of the dual can be larger than the cardinality of X
(ioo] see Exercise 12.32).
11.15 Let X be a Banach space of density character Ki that admits PRI.
Is it true that X* is not ^-separable?
Hint: No. ?i(c) has PRI and its dual is ^-separable. Indeed, ?\{c) C ^oo
(Exercise 5.50) and l^ is ^-separable; then use the restriction mapping.
11.16 (Kadec) Let X be a Banach space. Show that if X has an equivalent
Gateaux differentiate norm, then card(X) > dens(X*). Thus l^ has no
Gateaux differentiable norm.
Hint: Bishop-Phelps. card(/?N) > c.
11.17 Let X be a WCG Banach space. Show that C C Bx* is u>*-compact
if and only if C is ^-sequentially compact.
Hint: (Bx* , w*) is an Eberlein compact; Theorem 4.51 in co(r).
11.18 Let {xa] fa}aer be a Markushevich basis of a WCG Banach space
X. Show that card{(* G T; f(xa) / 0} < tt0 for every / G X*.
Hint: Let H = {/ G X*; card{a G T; f(xQ) ? 0} < tt0}- Clearly, H is a
closed subset of X* that contains all /a, and hence H is u;*-dense in X*.
By the Banach-Dieudonne theorem, we need only show that II D Bx* is
w*-closed in Bx*• Let g G H f) Bx* ¦ Since (Bx*,™*) is angelic, there
is a sequence hn ? H C\ Bx* such that hn ^U g. Since each ha has only
countable support over #a's, the same is true for g, meaning g G H O Px*.
11.19 Let X be a nonseparable WCG space. Show that there is a sequence
{xn} C 5x such that xn —> 0.
Hint: Let {x7} be a weakly compact Markushevich basis of X. There is
S > 0 such that ||x7|| > 5 for all 7 G T', where I" is uncountable. Then 0 is
a w-cluster point of x1\ 7 G T'. The result then follows from the angelicity
of w-compact sets.
11.20 Let X be a Banach space with dens(X) — Ni. Assume that {Pa; a <
uji} is a PRI in X. Show that given x G X) there is a < ui such that
* G Ptt(X).
Hint: dist(#, Pa(#)) is a nondecreasing function converging in a to 0, so it
must be eventually zero. Otherwise, taking an such that dist(a, Pan(X)) <
-, we would have a countable cofinal set in the segment [0,o;i]. Since uj\ is
the first uncountable ordinal, this is a contradiction.
11.21 Let X be a WCG space and {Pa; a < cji} be its PRI. Show that,
given f eX*, there is a < u>i such that / G P*(X*).

11. Weakly Compactly Generated Spaces 383
Hint: (Bx*,w*) is an Eberlein compact and thus an angelic space. Given
/ E Bx*, there is a sequence an < u\ such that P?n(f) ^ / in X*. Let
a < wi be the supremum of {an}. Then P* (X*) C P?(X*) for all n, so
11.22 Let X be a Banach space of density character Ki that admits a PRI.
Show that if cq is a subspace of X, then cq is complemented in X.
Hint: Each ? E Co lies in some Pa(X). Use the supremum of such a's. Then
use Sobczyk's theorem.
11.23 Let X be a Banach space of density character Ki with a PRI {Pa}-
Show that if the norm of X is Gateaux differentiate, then \jPa(X) = X
Hint: Given x E X, the function a —» ||# — P^(x)|| is continuous and equal
to zero at uj\. Thus it is zero starting from some countable ordinal.
Let / E Sx* attain its norm at x E Sx n Pa(X). Then ||P*(/)|| = 1
and P*(f)(x) — f(Pa(x)) = 1, and from the uniqueness of the support
functional due to Gateaux smoothness, we get P*(f) = /. If / E Sx* is a
general element, use the Bishop-Phelps theorem.
11.24 Let X be a Banach space. Show that if X* is not w*-separable, then
there is a subspace Y of X of density character &i that has a PRI.
Hint: Follow the proof of Lemma 6.15 with e = 0; use transfinite induction.
11.25 A Banach space X is called weakly countably determined (WCD) or
a Vasdk space if there exists a countable collection {Kn} of w*-compact
subsets of X** such that for every x E X and u E X** \ X there is no
for which x E ifno an(^ w ^ ^n0- Show that every WCG space is a Vasak
space.
Show that every closed subspace of a WCD space is a WCD space.
Hint: Let K be a ^-compact convex symmetric set generating X. Note
that X = (J nK. For n, m E N, put ifn>m = nif -f ^Bx** • Then /\n,m are
n
w*-compact in X**. Given x E X and u E X** \ X, choose m E N such
that dist(u,X) > ^ and n such that dist(;r, n/\) < ~.
11.26 Let X,Y be Banach spaces, T E #(X,Y). Show that if X has the
Grothendieck property and Y is WCG, then T is weakly compact.
Hint: By* is ^-sequentially compact since Y is WCG. Thus, T*(By*)
is iu*-sequentially compact and by the Grothendieck property w-compact.
Therefore, T* is weakly compact.
11.27 A Banach space X is said to have the Mazur property if
w*-sequentially continuous linear functionals are w*-continuous. Show that i^ does not
have the Mazur property.

384 11. Weakly Compactly Generated Spaces
Hint: t^ has the Grothendieck property. Every element of ?™ is
^-continuous and thus w*-sequentially continuous. If i^ had the Mazur property,
every element from ?™ would be it;*-continuous, and hence from ?oq. Thus
?oq would be reflexive, a contradiction.
11.28 ([FMZ]) Prove that if X is a Banach space such that for every norm-
ing subspace Y of X* there is an equivalent Frechet differentiable norm that
is Y-lower semicontinuous, then X is reflexive.
Hint: Use Exercise 3.43, the Smulian lemma, and the Bishop-Phelps
theorem.
11.29 Define T: Li[0,1] -+ C[0,1] by T(f): x ^ /* f(t) dt. Show that T is
continuous but not weakly compact. Show that T is completely continuous.
Hint: Take fn = 2n(x[i_i.ji] — X[i,i+i]) an<i show that {T(fn)} has no
weakly convergent subsequence.
To show that T is completely continuous, it is enough to show that T
takes tu-compact subsets of Li[0,l] to compact sets in C[0,1]. Prove this
using the Arzela-Ascoli criterion in C[0,1] and the Dunford criterion for
weak compactness in Li[0,1]: A subset K of Li[0,1] is relatively weakly
compact iff K is bounded and uniformly integrable, that is, for every e > 0
there is 8 > 0 such that fM \f\ dX < e for every / G K and M C [0,1]
satisfying X(M) < 6.
11.30 Consider the identity injection I from C[0,1] to L2[0,1]. Show that
it is continuous but not weakly compact.
Hint: L<i is reflexive whereas C[0,1] is not.
11.31 Consider <p E C[0,1]*, <p(f) = J01/2 f(t) dt - fi/2 f(t) dt. Show that
(p as an operator from C[0,1] to R is weakly compact but ^(Sc[o,i]) ls no^
w-compact in R.
Hint: Show that ^(?c[o,i]) = (—1,1).
11.32 Show that there is no one-to-one bounded linear operator from cq(T)
into any reflexive space if V is uncountable.
Hint: Assume that T is a one-to-one bounded linear operator from co(T)
into a reflexive space X. Then T is weakly compact, and hence so is
7~"\ Note that Bx* is ^-compact, T(Bx*) is compact in ^i(r), and
span™* (T(BX*)) = Mr). Hence c0(r)* is w*-separable, which is not the
case because all elements of co(r)* have countable supports.
11.33 Does there exist an isomorphic copy Z of ?2 in ?oq that is
complemented in ^oo?
Hint: No. ^ = C(K)} where K is the Stone-Cech compactification of N,
and thus ^00 has the Dunford-Pettis property. Then use Proposition 11.37.

11. Weakly Compactly Generated Spaces 385
11.34 Let p E [l,oo). Show that C[0,1]®?P is not isomorphic to C[0,1].
Hint: If p > 1, then ?p would be isomorphic to a complemented reflexive
subspace of C[0,1] and as such it would have the Dunford-Pettis property.
No infinite-dimensional reflexive space has the Dunford-Pettis property.
If p — 1, use the fact that C[0,1]* has an equivalent LUR norm ([Trol])
and ?oq does not have such a norm ([Lin7], [Trol]).
11.35 Show that lq[0,1] 0 ?2 is not isomorphic to Zq[0,1].
Hint: L\ is isometric to Loo, which is in turn isomorphic to t^
(Exercise 6.18), which is isomorphic to the space of continuous functions on the
Stone-Cech compactification of N, which has the Dunford-Pettis property.
Thus we get that L\ has the Dunford-Pettis property. Hence ?2 cannot be
isomorphic to a complemented subspace of L\ because it is reflexive and
infinite-dimensional.
11.36 Let p E A, 00). Show that Li[0,1] does not contain a complemented
subspace isomorphic to ?p.
Hint: Li[0,1] has the Dunford-Pettis property. Use Proposition 11.37.

12
Topics in Weak Topology
Theorem 12.1 (Eberlein, Smulian, Grothendieck; see, e.g., [Tod])
Let K be a compact set, and let L be a pointwise closed set in C(K).
Consider L with the pointwise topology of L C C(K). The following are
equivalent:
(i) L is compact.
(ii) L is sequentially compact.
(Hi) L is countably compact.
If L is, moreover, uniformly bounded in C(K), then (i)-(iii) are equivalent
to each of the following:
(i*) L is weakly compact.
(ii*) L is weakly sequentially compact.
(Hi*) L is weakly countably compact.
PROOF: (i) <=> (iii): This is the Grothendieck theorem; see [Tod].
(i) => (ii): By (i), L is angelic (see [Tod]), so (ii) holds.
(ii) => (iii) is straightforward.
In Theorem 4.51, we proved that (i*) to (iii*) are equivalent. To finish
the proof, we will show that in the case of a uniformly bounded pointwise
closed set L, (i) implies (ii*). We will use the following fact.
Claim
Let {fn} C L. Then there is a countable set D C K such that ifg,h G L
are in the pointwise closure of {fn} and g — h on D, then g — h.
Proof: Define the equivalent relation ~ on K by x ~ y if and only if
fn(x) — fn(y) for all n. Let S denote the collection of all equivalent classes

388 12. Topics in Weak Topology
x on K relative to ~. Call U C S open if and only if (p~l(U) is open in K,
where (p: K —> S is the quotient map x —>• x. Then 5 is a compact space
metrizable by the metric d(?,y) = J2^~n(fn(x) ~ /«(*/))> where the sum
runs through x G x,y G y.
Note that if g G L is such that g(x) = g(y) whenever fn(x) — fn(y) for
each n, then for x G 5 we can define g(x) = <7(#) for some # G ?• Then
g e C(S).
Let Dbea countable set in K such that <p(D) is dense in S. Then jD is the
desired set: Let g and h be in the pointwise closure of {/n} and g = h on D.
Let x, y G iv be such that # ~ y. Let /*, G {/n} be such that f„(x) —> ^(a:)
for every x E K. Then ^(y) = 1^G^(y)) = \im(fl/(x)) = g(x). Similarly,
we get h(x) — h(y). Hence we can define g and h by g{x) — g{x) and
h(x) — /i(x) for # G ?. Because D is dense in 5, we have g = h. Let x ? K
be given. Then <7(z) = g(x) = ft(f) = /i(ar), which proves the Claim.
We will now conclude the proof of the theorem by proving that, given
{/n} C Ly there is / G L and a subsequence {/nfc} of {fn} such that
Let D be as in the Claim. By the Cantor diagonal argument, let n\ <
ri2 < ... be such that lim(/nfc(x)) exists for every x G D. Let g and h
be accumulation points of {fn} in the pointwise topology (L is pointwise
compact). Then, by the choice of {n2}, g = h on D. By the Claim, g = h
on K. Hence {fnk} has a unique accumulation point and \im(fnk) = g in
the pointwise topology. Because L is uniformly bounded, by the Lebesgue
dominated convergence theorem, we get fnk —> g.
?
Eberlein Compacts
Corollary 12.2
A compact space L is an Eberlein compact if and only if there is a compact
space K such that L is homeomorphic to a pointwise compact subset of
C(K) considered in its pointwise topology.
Proof: Let L be an Eberlein compact. Then L is homeomorphic to a
weakly compact subset of some cq(T). Let K denote the one-point compact-
ification of the discrete space T. Then cq(Y) is isomorphic to a hyperplane of
C(K) by an isomorphism that is pointwise-to-pointwise continuous. Since
every weakly compact set in co(T) is pointwise compact and the weak and
pointwise topologies coincide on weak compact sets in co(r), we get that
L is homeomorphic to a pointwise compact set in C(K) considered in its
pointwise topology.

12. Topics in Weak Topology 389
On the other hand, let L be homeomorphic to a pointwise compact set
L\ in C(K) space for some compact space K. Let r be a homeomor-
phism of R onto (—1,-fl) and define a map $ from L\ into Bq{k) by
<$(/): x I—* r(f(x)). Then $ is a homeomorphism in the pointwise topology,
and <I>(Li) is thus a uniformly bounded, pointwise compact set in C(K).
By the preceding theorem, $(L\) is weakly compact. By Corollary 11.15,
<$(Li) is an Eberlein compact.
?
Proposition 12.3
Let K be a Hausdorff space. If M C C(K) is pointwise separable, then it
is separable.
Proof: ([Nam2]) Let D be a pointwise dense subset of M. Define (p: K —»
HD by <p(t): f —* f(t). Then (p is continuous and (p(K) is compact and
metrizable. For each / ? M, there is a unique / ? C(<p(K)) such that
/ = / o (p. Then (Af, || • ||) and (M, || -1|) are isometric. Since C((p(K)) is
separable, so are (Af, || ¦ ||) and (Af, || • ||).
D
Theorem 12.4 (Namioka; see, e.g., [Tod])
Let K be a compact set. Let L be a compact set in C(K) considered in the
pointwise topology. Then there is a dense G$ subset M of L such that the
pointwise and norm topologies ofC(K) coincide on M.
Proof: Unless stated otherwise, we will consider C(K) with its pointwise
topology.
Consider the (nonlinear) norm-norm-continuous map cp: C(K) —> C(K)
defined as (p(f): k h-> |- arctan(/(&)). Clearly, <p(C(K)) C Bq(k)- For every
S C C(K), the set S with the pointwise topology is homeomorphic via
(p with <p(S) C Bc(K) m the pointwise topology of C(K). (p is also a
homeomorphism of S in the norm topology onto <p(S) in the norm topology
of C(K). Thus, we may assume that the set L is bounded.
For / ? L, define the oscillation of the norm at / by
0(f) = inf{sup{||0i - g2\\] 9U92 G W); W C L, W open,/ ? W).
For n ? N, put Un = {/ ? L\ 0(f) < ^}. By the definition, it follows
that Un is an open set in L.
Let U — p| Un. Then U is a, Gs set in L. By the Baire category theorem,
the proof will be complete if we show that every Un is dense in L. Supposing
the contrary and replacing L by a nonempty set L\Un if needed, we can
assume that 0(f) > e > 0 for all / ? L and derive a contradiction. To this
end, we will construct a sequence {Vn} of open sets in L and fn ? Vn such
that Vn+i C Vn and dist(V^+i, conv{/i,..., /n}) > § for all n.
Assume that f\,..., fn and V\,..., Vn have been constructed. Put Kn =
conv{/i,..., fn}. Then Kn is a compact set in the norm topology inherited

390 12. Topics in Weak Topology
from C(K), so there is a finite f-j-net An C Kn. For each g ? An, let B(g)
be the closed ball in C(K) centered at g with radius ~. Note that B(g) is
pointwise closed in C(K).
Assume that B(g) contains an open set W in L. Then, for every gi and #2
in W, we have ||#i-#2|| < \\gi ~g\\ + \\9-~92W < ^ < e. Thus, every point
in W would have oscillation less than e, contradicting our assumption.
Hence B(g)f\L is closed and nowhere dense in L for every g ? An. There
is an open set Vn+i such that Vn+i C\ ( (J B(g)j = 0 and Vn+i C Vn.
KgeAn J_
Assume that dist(/, Kn) < e/3 for some / ? T4+i- Then dist(/, An) <
I + VI ~ ft• However, since / ? Vn+i and Vn+i fl (\JgeAnB(g)) = 0, we
get that dist(/, ^4n) > ||, a contradiction. Hence we may choose /n+i to
be any point in V^+i.
Let foo be an accumulation point of {/n}. Then /oo ? Vn for each n.
Because L is angelic, we get a subsequence {/nfc} of {fn} such that /^ =
lim(/nfc). Since L is bounded, we actually have fnk —> /oo, so by Mazur's
theorem, /^ ? conv{/nfc}. However, dist(/0O,conv{/ni,.. .,/«*}) > | for
every &, a contradiction.
?
Corollary 12.5 (Namioka [Naml], Benyamini, Rudin, Wage [BRW])
Lei L be an Eberlein compact. Then there is a dense Gs subset M of L
such that the topology on M inherited from L is metrizable.
PROOF: Let L be homeomorphic to a pointwise compact set L\ in C(K)
for some compact set K. By the preceding theorem, there is a dense Gs
subset M of L\ such that the pointwise and norm topologies from C(K)
on M coincide.
?
Corollary 12.6
Let L be an Eberlein compact. Then the set of all Gs points of L is dense
in L.
Proof: The result will follow from the following statement.
Fact 12.7
Let L be a compact set. If M is a metrizable set dense in L, then any point
of M is a Gs point of L.
Proof: Let On be open sets in M, f]On = {#}, and let On be open sets
in L such that On — On H M for n ? N. We will show that {On} forms
a neighborhood base at x in L. Indeed, let U be a neighborhood of x in
L. Find a neighborhood V of x in L such that V C U and choose n such
that On C V fl M. If z ? On and W is a neighborhood of z in L such that
VF C 6n, then there is some j/ ? MfW since M = L, so y ? OnflM = On.

12. Topics in Weak Topology 391
Hence On C On C V C U. This concludes the proof of the fact and of the
corollary as well.
?
Before stating another corollary, we need the following definition.
Definition 12.8
A topological space T is said to have property CCC (the countable chain
condition, or the Souslin property) ifT does not contain any uncountable
family of nonempty open pairwise disjoint sets.
Clearly, any separable space has property CCC. For metrizable spaces,
property CCC is equivalent to separability. The same applies to Eberlein
compacts.
Corollary 12.9 (Benyamini, Namioka, Rosenthal)
Let L be an Eberlein compact. If L is nonseparable, then L does not have
property CCC.
Proof: Let M be a dense metrizable subset of L. Then M is not separable
and thus there is an uncountable collection C of pairwise disjoint open sets
in M. For every C ? C, choose an open set C in L such that C — MCiC. We
note that if C\ and C2 are distinct members of C, then C1OC2 — 0- Indeed,
if x ? C\ 062 and U is a neighborhood of x in L such that x ? U C C\ 06*2,
then there is some y ? M f! U, so y E C\ 0 C2. Thus, {C\ C E C} forms an
uncountable pairwise disjoint collection of open sets in L.
?
Theorem 12.10 (see, e.g., [Eng])
Let X be a Banach space. Then X in the weak topology has property CCC
Proof: Let S C X* be an algebraic basis of X*. Then (X, w) is canonically
homeomorphic to a dense subspace of R5 endowed with the product
topology. It is enough to prove that R5 has CCC. To this end, let {Ui] i E /}
be a pairwise disjoint family of (basic) nonempty open sets in R5. Then,
for every i E I, there exists a finite set Si C S and a family {W%}S?S of
nonempty open subsets of R5 = R such that Ui — fj W%s and Wls = R for
all s? Si.
Let I0 C I be such that card(/0) < 2*° and set 50 = |J <%• Then
ieio
card(So) < 2«° and U{ = U W* x U R5 for i ? J0, so { fl W*}. ;
ses0 ses\s0 ses0 °
is a pairwise disjoint family of nonempty open sets in R5°.
We claim that R5° is separable; then Iq is countable and the proof is
finished. It is enough to prove that Q5° is separable and, since Q is
countable, it suffices to prove that Ns° is separable when N carries the discrete
topology. N5° can be identified with the space F of all mappings /: A —> N

392 12. Topics in Weak Topology
endowed with the pointwise topology, where A denotes the Cantor set. The
set A of all mappings /: A —* N that are constant on each of the 2n dyadic
subintervals of A at level n is countable and dense in F, so the separability
follows.
?
We quote one more lemma that will be used later; we refer to [Ros2] for
the proof.
Lemma 12.11 (Rosenthal)
Let T be topological space. If T has property CCC, then for every
uncountable family A of distinct open sets in T there exists an infinite sequence
oo
i*\, i<2,... of distinct members of A with f] F{, ^ 0.
«=i
Theorem 12.12 (Amir, Lindenstrauss [AmLi])
Let K be a compact space. The following are equivalent:
A) K is an Eberlein compact.
(ii) C(K) is weakly compactly generated.
(Hi) Bq(k)* i>n its w*-topology is an Eberlein compact.
(iv) There is a weakly compact set L C C(K) that separates points of K.
Proof: (i) => (ii): Assume without loss of generality that K is a weakly
compact set in c0(T) such that K C |i?Co(r)- Let $ be the family of all
finite sequences G1,...,jn) in T; note that 7; need not be distinct. For
n
(j) - G1,...,7n) G $ and x G K, define f^(x) = \[ x(ji). Then fa is
2 = 1
weakly continuous on K. Let A — {f<f>; (j) G $}U{1}. For every x G K
and e > 0 there are only finitely many distinct / G A such that \f(x)\ > e.
Indeed, if |/^(x)| > ^, then <f> is necessarily a sequence of length smaller
than n, and if |/#(e)| > e, then |arG»)| > e for all 7; G (/>. Therefore, every
sequence of distinct elements in A converges pointwise and thus weakly to
0, so AU {0} is weakly compact. Then span(A) is an algebra in C(K) that
separates points of K. By the Stone-Weierstrass theorem, span(-A) is dense
in C(K), showing that C(K) is weakly compactly generated.
(ii) => (iii): Theorem 11.16.
(iii) => (i): K is homeomorphic to a closed subset in Bc(k)* and ^nus
I\ is an Eberlein compact if Bc(K)* is.
(ii) => (iv): Any set S C C(K) for which spanE) = C(I\) necessarily
separates the points of K.
(iv) => (ii): Let L be a weakly compact set in C(K) that separates
points of K. For n G N, denote Ln = {/1 • • -/n; fc G L}, where /1 • • • fn
denotes the standard product of functions /i,...,/n- Every sequence in
Ln has a pointwise convergent subsequence in Ln. By Theorem 12.1, Ln
is a weakly compact set in C(K). For n G N, let sn — sup{||/||; / G Ln}.
Define ^ = (u^IB) U {1}.

12. Topics in Weak Topology 393
It is easy to show that A is weakly sequentially compact in C(K) (we
distinguish between the cases when a sequence lies in infinitely many -^-Ln,s
and when it does not). The subspace span(A) is an algebra that separates
points of K, and by the Stone-Weierstrass theorem, it is dense in C{K).
Thus, C(K) is weakly compactly generated.
?
Recall that a subset 5 of a compact space K is called a cozero set if there
is / E C(K) such that S = {k E K\ f(k) ^ 0}. A subset S of a compact
space K is a cozero set if and only if S is open and Fa.
A family of sets {Fa} is called point-finite in a set A if every point of A
lies in at most a finite number of sets Fa. It is called a-point-finite if it is
a countable union of point-finite families.
Theorem 12.13 (Rosenthal [Ros3])
A compact space K is an Eberlein compact if and only if it contains a a-
point-finite family T of cozero sets such that T weakly separates the points
of K; that is, given x\,X2 E K, x\ ^ X2, there is U E T such that either
x\ E U and X2 (fc U or X2 E U and x\ ? U.
Proof: Let T — \}Un be a family of cozero sets such that each Un is a
point-finite family of sets. Given n E N, for U E Un take /$ E C(K) such
that U = {x E K; ffi(z) ? 0} and 0 < /? < i. Put A = {/^; n E
N, J7 EWn}. It follows that every sequence of distinct elements in .A
converges pointwise and hence weakly to 0 (Lebesgue dominated convergence
theorem). The set A is thus weakly compact in C(K). Since A moreover
separates the points of K, K is an Eberlein compact by Theorem 12.12.
Conversely, assume that K is an Eberlein compact. By Theorem 12.12
and Theorem 11.12, let {fa]Fa}aqr be a weakly compact Markushevich
basis of C(K). Assume without loss of generality that ||/aJ| < 1 for every
a E T. Define cozero sets U^j = {x E K\ ^ < fa(x) < ^} for a E T,
3 € Z, \j\ > 2. For n E N, put
Hn = {U^\ ^Er,iEZ,2<|i|<n}.
Note that, for every infinite sequence of distinct o^-, we have fai —» 0
pointwise and that (J U?j = {x E K\ |/ct(^)| > ~}• Thus Un is point-
2<|j|<n
finite. If ?i,#2 E X, x\ ^ a?2, then fa(xi) / fa(x2) for some aGT and
then J7? • separates x\ and ar2 for some j and n.
D
To compare Eberlein compacts with metrizable compacts, we state the
following result, whose proof we omit.
Theorem 12.14 (see [Ros3])
A compact space K is metrizable if and only if it contains a a-point-finite
family T of cozero sets that separates the points of K in the sense that for
every x\,X2 E K, x\ / X2 there is G E T such that x\ E G and X2 ? G.

394 12. Topics in Weak Topology
Uniform Eberlein Compacts
Definition 12.15
A compact space K is called a uniform Eberlein compact if K is
homeomorphic to a weakly compact subset of a Hilbert space in its weak topology.
Theorem 12.16 (Benyamini, Starbird [BeSt])
Let K be a compact space. The following are equivalent:
(i) K is a uniform Eberlein compact.
(ii) K is homeomorphic to a weakly compact set K in co(T) which has
the property that for every e > 0, there is N(e) G N such that card{7 G
T; |*(t)| >e}< N(e) for all k G K.
Proof: (i) => (ii): Let K be homeomorphic to a weakly compact set K
in ^(r) for some I\ Let K C mBi2(r) for some m G N. Given e > 0 and
k G K, if S denotes all those 7 G T for which \k(j)\ > e, then we have
m2 >X^lfe(T)|2 > cardE)^2.
5
The implication now follows because the formal identity map from ^(T)
into co(r) restricted to K is weak-weak continuous.
(ii) => (i): Assume without loss of generality that ||fc|| < 1 for every
k G K. Let /:[—1,1] —> [—1,1] be a continuous, strictly increasing, and
odd function such that /(?) < Bn^Dr))~^ for all n G N. Define a
map (p from K into l^(T) by (p(k): 7 h-» f(k(j)). Given k G K and n G N,
put An = {7 6 r; ^ < \k(j)\ < 1}. Then
\Hmi(v) = Ei/(*mi2 = EE i/(^))i2
7 n j?An
n n
n n
Thus y? maps K into ?^2(r), and it is clearly one-to-one and
continuous in pointwise topologies of cq(T) and ^(r), which coincide with weak
topologies on bounded sets in these spaces.
?
Theorem 12.17 (Benyamini, Starbird [BeSt])
Let K be a compact space. The following are equivalent:
(i) K is a uniform Eberlein compact.
(ii) There is a Hilbert space H and a bounded linear operator T from H
into C(K) such that T(H) is dense in C(K).
(Hi) (Bc(k)* ,w*) is a uniform Eberlein compact.

12. Topics in Weak Topology 395
Proof: (i) => (ii): Similarly as in the proof of (ii) => (i) in
Theorem 12.16, we show that K is homeomorphic to a weakly compact subset
K of ^(r) for some Y such that 0 ? K and J^ |#(t)| < 1 for every x E K.
oo
Set A0 = 0 and for n G N, put ,4n = T x • • • x T (n times), let A - |J Anj
n=zO
and H — [J2^(^)) . Define an operator T from i7 into C(K) as follows.
For h = (h,A)AeA G # and x G iv, we put
n A=(a1,...,an)eAni=l
where x is considered an element of ^(r) = ^(r)* and {e7} is the canonical
basis of ^2 (r).
Then T is a bounded linear operator because for every x G K and \\h\\ < 1
we have |a:(/i^)| < ||/&a|| < |H| < 1 for all A and thus, since J2 \x(l)\ < 1>
7Gr
we have
n n
? Y[x(ai)x(hA)\ < Yl IIl^^.)!
A=(ofi,...,an)€-4n * = 1 A = (ail...>orn)e-4n * = 1
= (^We7)|)n<l.
The set TG7) contains all polynomials in the coordinate functionals
x(ea). These polynomials separate points in K and do not have a
common zero because 0 ^ K. Therefore, T(H) is dense in C(K) by the
Stone-Weierstrass theorem.
(ii) => (iii): The dual operator T* of the operator from (ii) maps C(K)*
one-to-one and w*-w*-continuously into H* — H. Therefore, BC(k)* is
homeomorphic to a weakly compact subset of a Hilbert space.
(iii) =4> (i): This follows from the fact that K is w*-closed in Bc(k)* •
?
The following result was motivated by [ArFa], [Far], [MMOT], and [Tro2].
Theorem 12.18 ([FGZ])
(i) A Banach space X admits an equivalent UG-smooth norm if and only
if (Bx* ,w*) is a uniform Eberlein compact.
(ii) Let K be a compact space. C(K) admits an equivalent UG-smooth norm
if and only if K is a uniform Eberlein compact.
The proof is based on the following result.
Lemma 12.19
Let X be a Banach space. If X admits an equivalent UG-smooth norm,

396 12. Topics in Weak Topology
then X is a Ka$ subset 0/(X**, iu*); that is, X = f] (J Km^n, where
n>l ra>l
Km,n are w*-compact sets in X**.
Proof: Let || • || be an arbitrary equivalent norm on X. For n,p ? N, we
define subsets 5n>p(|| • ||) of X as follows.
SnA\\'\\)= {x?X\ \f(x)-g(x)\<±
for all f,ge Bx* with ||/ + flf|| > 2 - ?}.
Now consider an equivalent UG-smooth norm || • || on X. Its dual norm
is W*UR; that is, fn — gn —> 0 whenever {/n} and {#„} are bounded in
X* and 2||/n||2 + 2||yn||2 - ||/n + gn\\2 -> 0 (see the proof of Lemma 8.4 or
Theorem II.6.7 of [DGZ3]). Thus Q Sn,P(\\ -\\) = x for every P € N. We
n>l
will show that X = pi U SnA\\ ¦ llf •
p>ln>l
By the last equality, it suffices to prove that for any G ? X** \X there
is p ? N such that
G^U^dl-H)'0"- (*)
n>l
Denote H ~ G_1@) C X*, and recall that it is a norming subspace of X
(Exercise 3.43). Define an equivalent norm q on X by
g(x) = sup{|/(*)|; feBH}-
We claim that 5n,p(|| • ||) C Sn#(q). Observe that, by the bipolar theorem,
the dual unit ball Bq* satisfies
?,. ={/€**; «*(/)< 1} = 5T*. (**)
Therefore, if q*(f) < 1, q*(g) < 1, and q*(f + g) > 2 - 2/n, there are
nets {fa} and {#a} in iJ w*-convergent to / and g, respectively, such
that ||/a|| < 1 and \\ga\\ < 1 f°r all <*• Since the norm g* is u>*-lower
semicontinuous, q*(fa + <7a) > 2 — \ when a is large enough. Because || • ||
and q* coincide on H, \\fa + ga\\ > 2 — 2/n for large a. If x ? 5n>p(|| • ||),
we have |/a(#) — #<*(#) | < 1/p f°r <* large enough, and hence z ? SniP(q).
This shows our claim.
Thus, to prove (*) it is enough to show that G ? (J SnjP(q) for some
n>\
p large enough. To this end, choose p ? N such that p > l/</**(G). Fix
n ? N and for simplicity set 5 = ?Vi ,?(<?)• We must show that G ? S .
Choose / ? Bq* such that G(f) > l/p and find x ? X such that #(#) < 1
and/(a:) > l-?. By (**), there is g ? # withg*(#) < 1 and g(x) > l-?.
We then have ?*(/ + flf) > (/ + </)(*) > 2 - ?.

12. Topics in Weak Topology 397
By the definition of S, for all z G S one has (/ — g){z) < \- Thus, if
G G S then G(f — g) < -. This contradicts the fact that G(f — g) =
G(f) > }•
For every pEN, let the family {Kp,q}q be equal to the reindexed family
{Snj) C\rnBx**}n,m' Then X is Kas in X** in its w*-topology.
D
Proof of Theorem 12.18: (Sketch) If (Bx*,w*) is a uniform Eberlein
compact, then by Theorem 12.17 there is a bounded linear operator T
from a Hilbert space H onto a dense set in C(Bx*)- By Theorem II.6.8
of [DGZ3], C{Bx*) admits an equivalent uniformly Gateaux differentiable
norm and so does its subspace X.
If X admits an equivalent uniformly Gateaux differentiable norm, then
X is K<r8 in -^** in its w*-topology by Lemma 12.19 and thus it admits
projectional resolutions of the identity (see Theorem VI.2.5 of [DGZ3]).
Then the Rosenthal criterion (see Theorem 12.13) can be used to prove
that Bx* is a uniform Eberlein compact ([FHZ]).
?
Corollary 12.20
Let X be a Banach space. If X admits an equivalent UG-smooth norm,
then w*-dens(X*) = dens(X).
For the proof, see Corollary 12.52 and Chapter VI of [DGZ3].
We remark that a Banach space X admits an equivalent UG-smooth
norm if X admits a UG bump ([Tan]).
Definition 12.21
The norm || • || of a Banach space X is called weakly uniformly rotund
(WUR) if (xn — yn) —» 0 whenever xn G Sx and yn G Sx are such that
\\zn + yn\\ ->2.
The following result follows from Corollary 12.20. However, we will give
a direct proof of it here.
Theorem 12.22 ([Haj2])
If the norm of a Banach space X is WUR, then X is an Asplund space.
Proof: (Godefroy) By Theorem 8.26, it is enough to show that if the norm
of a separable Banach space X is WUR, then X* is separable.
For n G N, put
Vn = {fe Bx*] \f{x-y)\ < I for x,y G Bx such that ||x + y|| > 2- ?}.
Note that Bx* =\JVn-
Since (Bx*, w*) is a metric compact space, for every n G N there is a
countable w*-dense set Sn in Vn. We claim that span{(j5n} = X*.

398 12. Topics in Weak Topology
Assume that this is not the case and find F G Sx** such that F(f) — 0
for all / G \JSn- Choose /0 G Sx* such that F(fo) > §, and assume that
/o G V^q. Choose {xa} G #x such that xa -* F (Goldstine). By the w*-
lower semicontinuity of the second dual norm, assume that ao is such that
\\xa -f- xp\\ > 2 — ~ for every a,/? > c*o, and hence \f(xa) — f(xp)\ < | for
all a,/? > a0 and / eVno.
Since xa ^ F, we have \F{f) - /(a?tt0)| < | for all / G Vno. Thus, for
f ? Sno) we have
l(/-/o)(*a0)| = |F(/)-Fao) + /(^0)-F(/) + F(/o)-/o(^0)l
> |F(/) - F(/0)| - |/(xao) - F(/)| - |F(/0) - /0(*ao)|
> 8 _ 6 _ 2
— 9 9 ~~ 9'
which contradicts the fact that /o G Sno
?
Definition 12.23
A compact space K is called a scattered compact if every closed subset
L C K has an isolated point in L.
Recall that a point p is isolated in K if there is a neighborhood U of p
in K such that U C\ K — {p}. Note that a compact space K is scattered if
and only if every subset of K has a relatively isolated point.
We will consider the Cantor derivative of a set K: K^ — K, K^ — K',
the set of all cluster points of K (i.e., the points that are not isolated). If a
is an ordinal and K^ are defined for all /? < a, then we put K^ = (K^)'
for a = /? + 1 and #(") = f| K^ for a limit ordinal a.
A compact set K is scattered if and only if KW = 0 for some ordinal 7.
Indeed, if K is not scattered, there is a nonempty perfect set L C K (i.e.,
a closed set all points of which are cluster points of L). Then Lcf| K^
a
and thus K^ ? 0 for all a. If L = f]K^ ^ 0, then L is a perfect set in
a
K, so K is not scattered by the definition.
If K is scattered and an ordinal A is the first ordinal such that K^ — 0,
then A is not a limit ordinal. Indeed, by the compactness of K, we have
that K^ = p| KW ^ 0 for all limit ordinals. Thus, A = /? + 1 and p is
p<\
the last ordinal such that K^ / 0. By the compactness of K^\ we get
that then K^ is a finite set in K.
Lemma 12.24
A continuous image of a scattered compact is a scattered compact.
Proof: Let A be a scattered compact and let / be a continuous map
of K onto a compact space L. Assume that P is a perfect subset of L.

12. Topics in Weak Topology 399
Consider the family A of all compact subsets A of K such that f(A) — P
ordered by inclusion. If {Aa} is a chain in A, then by the compactness,
f]Aa ^ 0, and also f(f]Aa) = P. Let B be a minimal set in A. Since K
is scattered and B is compact, B contains a point q that is isolated in B.
Put B' — B\ {q}. Then Bl is compact and the minimality of B gives that
f(B') is a proper subset of P. Note that f(q) (? f(B') because otherwise
f(B') — f(B) — P. Since f(B') is compact, f(q) is not a cluster point of
f(Bf). Since P \ f(B') — /(<?), we get that f(q) is not a cluster point of
P either. Therefore, the set P contains a point f(q) that is isolated in P.
This shows that P is not perfect, a contradiction.
?
Lemma 12.25
A compact space K is countable if and only if K is metrizable and scattered.
Proof: Assume that K is countable. For every (x,y) E K x K with x / y,
the set f/^y = {/ E C(K); f(x) = /(y)} is a closed hyperplane in C(K).
By the Baire category theorem, there exists / E C'(A') that belongs to
no HX)V, and hence / is a one-to-one map of K into R. Therefore K is
metrizable. Any closed subset L of K is countable and has isolated points
in L by the Baire category theorem. Hence K is scattered.
Conversely, assume that K is metrizable and scattered. Because the
topology of K has a countable basis, it follows that there is a countable
ordinal c*o such that /\(a°) = 0. Every set in K is separable and thus
K(a) y A-(or+i) is countable for any a. Hence K is countable.
?
Corollary 12.26
Let K be a scattered compact. If f is a continuous map from K into R,
then f{K) is countable.
Proof: f(K) is metrizable and scattered.
?
Lemma 12.27 (Rudin; see, e.g., [DGZ3])
Let ii be a nonnegative regular finite Borel measure on a scattered compact
space K. If fi({p}) — 0 for every p E K, then fi vanishes identically on K.
Proof: Suppose fJ>(K) > 0. Let an ordinal A be the last ordinal such that
I{W zfz 0. Then K^ is a finite set. Thus ^(KW) = 0 by the assumption on
/i. Let a be the first ordinal such that fi(K^) < n{K)- If a = /3+1 for some
/?, then fi(KW) = fi(K^) + fi(K^\K^). The set K^\K^ contains
no infinite compact set by the definition of the Cantor derivative. Hence, by
the regularity of// and the property that /i vanishes on all singletons, we get
that fi(K^\KW) = 0. Thus fi(K^) = fi(K^), contradicting our choice
of a. Hence a is a limit ordinal. By a simple compactness argument, for

400 12. Topics in Weak Topology
every open set G that contains K^a\ there then exists an ordinal /? < a such
that KM C G. Since /? < a, we have fi(K^) = fi(K). Hence /i(G) = fi(K)
for every open set G containing I\(a\ From the regularity of /i we obtain
fi(K^) = fi(K), a contradiction.
n
Theorem 12.28 (Rudin [Rudl])
//A' 25 a scattered compact, then C(K)* is isometric to ^i(T) for some I\
Proof: Let // be a nonnegative regular finite Borel measure on a scattered
space K. Let 5 be a collection of all points p of K satisfying p>({p}) > 0.
Note that {p E K] /i({p}) > s} is finite for every e > 0 since // is a finite
measure. Thus 5 is countable. Define a measure v on Borel subsets A of K
by z/(yt) = //(AnS'). By Lemma 12.27, the measure fi — v vanishes on K and
thus \i — v. Since S = {qn} is countable, by the general measure theory,
oo
JKfdfi = $Kfdv - J2 cnf(qn) for every / G C(K), where ^, \cn\ < oo.
z = l
Conversely, given {cn} with ^ |cn| < oo and {qn} C i^, the functional F
defined for / G C{K) by F(f) = Y1 cnf(<ln) is a continuous linear functional
on C{K). Clearly, ||F|| < J2 \cn\- To get the opposite inequality, for a finite
set {gi,..., qn] we consider / G BC(K) such that f(qn) = sign (c„).
D
Note that the isometry in Theorem 12.28 is not in general a dual map.
Indeed, cq is not isometric to c (Exercise 3.75), yet cj is isometric to c*
(Exercise 2.24).
Theorem 12.29
Let K be a compact space. K is a scattered compact if and only if C(K) is
an Asplund space.
Proof: Assume that K is scattered and X is a separable closed sub-
space of C(K). Choose a dense sequence {#n} in Bc(K) and define a map
G: K -> [-1,1]N by G(k) = {#n(fc)}. The compact L = G(K) is scattered
by Lemma 12.24 and metrizable. Thus L is countable by Lemma 12.25, and
hence C(L)* is isometric to l\ by the proof of Theorem 12.28. Therefore
C(L)* is separable. Moreover, X is isometric to a closed subspace of C(L)
and X* is thus separable.
Conversely, assume that K is not scattered and C(K) is an Asplund
space. Let P be a perfect subset of K. The restriction of continuous
functions on K to P is a bounded linear operator from C(K) into C(P), which
is onto due to the Tietze extension theorem. Therefore, C(P) is a quotient
of an Asplund space. If Z is a separable closed subspace of C(P)} then there
is a separable closed subspace W of C(K) that is mapped onto Z by the
quotient map (Exercise 2.47). Thus Z* is isomorphic to a closed subspace
of W* (Exercise 2.39). Since W* is separable, we have that Z* is
separable. Hence C(P) is an Asplund space. In Exercise 8.28, we proved that the

12. Topics in Weak Topology 401
supremum norm in C[0,1] has no point of Frechet smoothness. Since P is
perfect, we similarly prove that the supremum norm of C(P) (and hence
of Z) has no point of Frechet smoothness. This contradicts Theorem 8.26.
?
The following theorem lists some topological properties of compact sets
and C(K) spaces.
Theorem 12.30
Let K be an infinite compact space.
(i) C(K) contains an isomorphic copy of Cq.
(ii) K has property CCC if and only if every weakly compact set in C(K)
is separable if and only if C(K) does not contain any isomorphic copy of
Co(r); T uncountable.
(Hi) If K contains a nontrivial infinite convergent sequence, then C(K)
has a quotient isomorphic to Cq.
(iv) If K is scattered, then C(K) contains an isomorphic copy of c$ that is
complemented in C(K).
(v) K is scattered if and only if C(K) is saturated with isomorphic copies
of c0.
Proof: (Sketch) (i): Let {Un} be an infinite sequence of open pairwise
disjoint sets in K. For each n, let fn G C(K) be such that fn(tn) = 1 for
some tn G Un> 0 < fn < 1, and fn = 0 outside Un- Then consider the
operator T: c0 —> C(K) defined by T((an)) = ^anfn.
n
(ii): Assume that S is a nonseparable w-compact symmetric convex
subset of C(K). As in Theorem 11.12, we find an uncountable subset M of
5\{0} such that every sequence of distinct members of M weakly converges
to 0 in C(K). Thus there isj) > 0 such that M = {g G M; \\g\\ > 6} is
uncountable. For every g G M, let Ug = {t G K\ \\g(t)\\ > 8/2}.
Let <7i,02>--- be an infinite sequence in M. Then {U9i} is an infinite
oo
sequence of open nonempty sets in K. If t0 G f] U9l, then |<7;(?o)| > f for
2 = 1
OO
all i, a contradiction with gi —» 0. Hence p| U9i — 0. Then K does not
i=i
have property CCC due to Lemma 12.11.
Thus, we showed that K does not have property CCC if C(K) contains a
^-compact nonseparable subset. To finish the proof of (ii), we now assume
that for some uncountable T, co(r) is isomorphic to a subspace of C(K).
Let {e7} be the canonical basis of cq(T). Then {e7}u{0}isa nonseparable
and ^-compact set in C(K).
Finally, if K does not have property CCC, then similarly to (i), C(K)
has a subspace isomorphic to co(T) for some uncountable T.
(iii): Let an G K be distinct with lim(an) = a G K. Then {an} U {0} is
a compact subset of K and C({an} U {a}) is isomorphic to c, which is in

402 12. Topics in Weak Topology
turn isomorphic to Co (Exercise 5.32). Then we can use Tietze's extension
theorem to show that the restriction of elements of C(K) to {an} U {a} is
a bounded linear operator of C(K) onto C({an} U {a}).
(iv): Let {an} be an infinite sequence of isolated points of K. Since K is
scattered, C(K) is an Asplund space (Theorem 12.29) and thus Bc(k)* is
^-sequentially compact (see the proof of Theorem 10.10). Hence we can
assume that an —> a E K for some a. Given / E C(/\), put f(t) — f(a) if
t zfz an for all n and f(t) — f(an) ift = an. Then / E C(K), the operator
P:C(K) -> C(tf) denned by P(f) = / is a projection, and P{C(K)) is
isomorphic to c, which is in turn isomorphic to Co- Since Co is not isomorphic
to any quotient of ^oo (Exercise 6.31), we obtain as a corollary that /?N
does not contain any nontrivial convergent sequence.
(v): If K is not scattered, K contains a perfect set K and thus there is a
continuous map (p of K onto [0,1] (see, e.g., [Lac2]). By Tietze's extension
theorem, there is a continuous map <poiK onto [0,1]. Using the composition
map, we see that C[0,1] is isomorphic to a subspace of C(K), and C[0,1]
contains for instance ?2 which does not contain c$. UK is scattered, we refer
to [Lacl] for the proof that C(K) is saturated with subspaces isomorphic
to Co-
?
Recall that a compact K is called totally disconnected if for every x /
y E K there exists a clopen set U such that x E U and y (? U. It is an
open problem whether for every compact K there is a totally disconnected
compact K\ such that C{K\) is isomorphic C{K). This problem has a
positive solution for metrizable compacts ([PeBe]).
Weakly Lindelof Spaces
Theorem 12.31
Let A be a set. For a E A, let Xa be separable topological spaces, and let
Y be a topological space all points of which are G5. For every continuous
function /: f| Xa —+ Y, there is a countable subset S C A and a continu-
ous function fs: f| Xa —> Y such that f — fs ° Ps, where ps denotes the
canonical projection of Yl Xa onto Yl Xa.
a?A aeS
Proof: Let x E X — Yl Xa- Let y — f(x). Since {y} is a Gs set,
aeA
there exists a sequence {V(y, n)}n<L1 of open subsets of Y such that
00 00
{y} = PI V(y,n). Then f~\y) = f| Z'1^(!>>")]• Let U{x,n) be a
n=l n=l
basic open subset of X such that x E U{x,n) C f~1[V(y} n)], n E N.

12. Topics in Weak Topology 403
oo
Then x G Wx — f] U{x,n) C f~1(y)- Given a basic open set U C X,
n = l
define S[U] = {a G A; pa(U) ^ Xa} (a finite subset of A) and let
oo
Sx = (J ?[[/(?, n)]. Observe that /(x) = f(x') if #,#' G X satisfy
n = l
Psx(x) = Psx(x')-
Fix a ? X and set aa = pa(a) for a G A. Given a countable subset
5 of A, we define the mapping e: fj Xa —> Yi Xa so that, for every
aeS cc?A
/i6 f] Xa, e(h) is the following "extension" of h:
exes
r P5(c(A)) = fc,
\ pa(e(h)) = aa,a G -A \ 5.
Set 5o = Sa- By induction, assume that So C Si C ... C Sn have been
defined. Let Dn be a countable dense subset of Yl Xa- Define Sn+i =
aesn
Sn U (J 5e(dn). Note that all Sn are countable subsets of A. Finally, put
OO OO
S = (J Sn and L> = {a} U \J e(Dn). It is easy to see that E = ps(-D) is
n=0 n=0
dense in J\ Xa.
a?S
Claim
Ifx,yeX satisfy ps(x) = ps(y), ihen f(x) = /(t/).
Proof: It is enough to check f(x) = f(e(ps(x))) for all xGl.To this end,
first note that f(ps\h)) = /(e(ft)) for all /i G ?. This is a consequence of
the following observation: if x G p^fo), then pse(k) (e(^)) = jP5c(fe)(^)«
Now, given h E E and xGl, define Xh G X so that
f Ps(*a) = ft,
\ PA\s(vh) = *¦
We have /(a?^) = /(e(A)). Since i? is dense in ]~J Xa, there exists a
net {di} in D such that ps(di) —> Ps(aO- It follows that #Ps(<ft) —» a?
in X. Since / is a continuous function on X, f(xps(di)) ~+ f(x)- We
also have e(xps(di)) -> e(ps(x))y and hence f(xps(dt)) = /(e(arps(di))) ->
/(e(ps (#))), which proves the claim.
Thus we can define fs: n ^<* ""* ^ sucn tnat f — fs °Ps- Clearly, /s
is continuous.
?
Theorem 12.32
Let X be a Banach space, and let W be the family of all real-valued functions
on X that are w-continuous. Then card(PV) = card(X*).

404 12. Topics in Weak Topology
Proof: Let A be an algebraic basis of X*. We may assume that X is
infinite-dimensional, so card(yt) > 2N (Exercise 6.4). Since (X, w) can be
canonically identified with a dense subspace of R/4 endowed with the
product topology, there are as many ^-continuous functions on X as continuous
functions on HA. By Theorem 12.31, given a real-valued continuous
function / on HA, there exists a countable subset Aq of A and a factorization
/ = foopAo, where /o: RA° —* R is also continuous. Since RAo is separable,
there are 2^° real continuous functions on it. We have card(A) = card(X*),
so the number of countable subsets of A is also card(X*). The statement
now follows.
?
Recall that a topological space T is called Lindelof if every open cover
of T has a countable sub cover. We will show (Theorem 12.35) that every
WCG space is Lindelof in its weak topology (we say weakly Lindelof).
Theorem 12.33 (Orihuela [Ori])
Let X be a Banach space. Assume that for every map <j> from X into finite
subsets ofX*, X admits a norm-one projection P of X onto P(X) such that
P(X) is separable and for some countable dense set A C P{X), P*(f) — f
for all f G <t>(A). Then X is weakly Lindelof.
Proof: Consider X in its weak topology. Assume that {K*}aer is an open
cover of X. For every x G X, let rx > 0 be the supremum of all positive
numbers r such that B%(#, r), the open ball of radius r centered at x) lies
in some Va. Choose Vx G {VQ} so that B^(x)rf) C Vx and assume that
Vx is formed by the intersection of half spaces given by a finite set Kx of
functional. Define <j)(x) = Kx.
By our assumption, there is a projection P of X onto P(X) and an
appropriate set A C P{X) constructed for the map <j>. We claim that {Vz; z G A}
is a cover of X, which will complete the proof.
Choose any x G X and let B^(P(x)Jr) C Vp(x). Find z G A such that
z G B%(P(x),±r). Then rz > ^r, and thus B%(z,\r) C Vz. Hence
P(x) G Vz. Since </>(z) C P*(X*), it follows that x G Vz. Indeed, if, say,
\f(Px - z)\ < e for all / G <f>(z), then \f(x - z)\ = \P*(f)(x - z)\ -
\f(P(x) - P(z))\ = \f(P(x) -z)\<e. This shows that {Vz} is a cover of
X.
?
Theorem 12.34 (Amir, Lindenstrauss [AmLi])
Let X be a weakly compactly generated space. If <j> is a map that assigns
to each x G X a finite set <f)(x) G X*, then there is a norm-one projection
P of X onto P(X) such that P(X) is separable, and there is a countable
dense set A C P(X) such that P*(f) = / for all f G ^(A).

12. Topics in Weak Topology 405
Proof: Similar to the beginning of the proof of Lemma 11.10. The only
adjustment needed is that we play the "exhaustion game" also in X.
?
From Theorems 12.33 and 12.34, we obtain the following theorem.
Theorem 12.35 (Preiss, Talagrand [Tall])
Every weakly compactly generated space is weakly Lindelof.
A Banach space in its weak topology is Lindelof if and only if it is
paracompact if and only if it is normal ([Bau], [Rez]).
We will see that the Lindelof property is not a three-space property. This
is not the case with the following (weaker) property, which was introduced
by Corson in [Cor].
Definition 12.36
Let X be a Banach space. A closed convex subset M of X is said to have
property C if for every family A of closed convex subsets of M with empty
intersection there is a countable subfamily B of A with empty intersection.
We say that X has property C if the set X has property C.
In other words, X has property C if and only if every family T of
complements of closed convex sets in X that covers X has a countable subfamily
that covers X. Hence, every X that is weakly Lindelof has property C.
Note that property C passes to closed subspaces and quotients.
Theorem 12.37 (Pol [Pol2], [Pol3])
Let Y be a closed subspace of a Banach space X. If both Y and X/Y have
property C, then X has property C.
In the proof, we will use the following statements.
Lemma 12.38
A Banach space X has property C if the following condition is true:
(*) If a family K of nonempty closed convex sets in X is closed under count-
able intersections, then for every a > 0 there is a ? X with dist(a, C) < cr
for every C G /C.
Proof: Let C be a family of nonempty closed convex subsets of X closed
under countable intersections. We must show that f]C ^ 0.
To this end, define inductively a sequence a« G X and nonempty closed
convex sets Cl for every C G C in such a way that:
A) C = C°DC1D.C2..., diam(Ci+1) < 2"';
B) every collection Cl — {C%\ C G C} is closed under countable
intersections;
C) dist(a?, Cl) < 2--1, i = 0,1,2,....
We choose ao by using (*) with K — C and a — 2_1. Assume that
0 and cij are chosen, j = 0,1,2,..., i. For every C G C, put Cz+1 =
Cl fl (cii + 2~z~1Bx)- The sets C*+1 are closed, convex, and nonempty by

406 12. Topics in Weak Topology
C), the condition A) is satisfied, and the family Ci+1 = {Ci+1; C G C}
satisfies B) as Cl does. We complete the inductive step by choosing a2+i
by applying (*) to K — Cl+1 and a ~ 2~l~2.
The points a; form a Cauchy sequence, because by C) and A) we have
Wat - ai+1\\ < 2'-1 + 2"'+1 + 2'-2 < 3 • 2~i+1
for i > 1. The limit point of {a2} yields the conclusion.
?
Lemma 12.39
// a Banach space X does not have property C, then there exists e > 0 and
a family C of nonempty closed convex subsets of Bx closed under countable
intersections such that for every closed convex subset M of X with property
C there is a Cm G C with dist(M, Cm) > ?-
Proof: Because X does not have property C, by Lemma 12.38 the property
(*) fails for some K and some a > 0. Since /C is closed under countable
intersections, there is a natural number n such that each member of K
intersects the ball nBX- Put C = {~{C n nBx)\ C G K,} and e - %. The
family C consists of nonempty closed convex subsets of Bx, it is closed
under countable intersections, and for every x G X there is Cx G C with
dist(x, Cx) > e.
Let M be a closed convex subset of X with property C and suppose by
contradiction that dist(M, C) < e for C G C. Put C" = (C + sBx) H M for
every C G C. The sets C" are closed, convex, and nonempty.
Given a countable collection A C C, we have (H^)' C Hi^'j C G ^1}
(note that f|^4 G C). Hence, by property C of M, there is x G C\{C'\ C G
C}. But then dist(#, C) < e for every C G C, a contradiction.
D
Proof of Theorem 12.37: Let g be the quotient map of X onto X/Y.
Assume that X does not have property C. Find e > 0 and C by Lemma 12.39.
Since for every z G X/Y', tf^) has property C (because Y has property
C), there is a set Cz G C with dist^'1^), C*) > ?. Thus z (? q(Cz), and
hence f]{q(C)] C G C} = 0. This is a contradiction with property C of
X/Y.
?
The following result is from [Cor].
Theorem 12.40 (Corson [Cor])
The space C[0,c*;i] does not have property C.
PROOF: Observe that the hyperplane L — {x G C[0,u;i]; #@) = 0} of
C[0,cji] is isomorphic to C[0,u;i]. Since all hyperplanes of a given
Banach space are isomorphic (Exercise 2.7), the hyperplane Co[0,u>i] — {x G
C[0,o;i]; x(u>i) = 0} is isomorphic to C[0,o;i]. Therefore, it suffices to show

12. Topics in Weak Topology 407
that Co[0,o;i] does not have property C. For a <u>i, define xa G Co[0,u>i]
by
c(8)-f° if^>Q;>
For a € [0,wi), let Ka = {x G Co[0,ui]; \\x - xa\\ < \}. If a,- < wx for
OO OO
i G N and sup (a,-) < /? < uu then |ar/j G f| ^- Hence f| #ai ^ 0. If
*=i i=i
x G p| Ka, then #(a) = | for all 0 < a < o;i, which is impossible by the
arOi
definition of Co[0, wj. Thus f| Ka = 0.
D
Theorem 12.41 (Pol [Pol2])
yl Banach space has property C if and only if for every A C Bx* and
f G A there is a countable subset B of A such that f G conv™ (B).
In the proof, we will use the following lemma.
Lemma 12.42
Assume that a Banach space X does not have property C. Then there is
A C Bx* and e > 0 such that:
A) for every closed subspace M of X with property C, there is x* G A
vanishing on M;
B) for every countable B C A, there is x G Bx with x*(x) > e for all
z*G?.
Proof: Find C and e > 0 by Lemma 12.39. Let M be the family of all
closed subspaces of X with property C. Given M G M, by Lemma 12.39
there is Cm ? C such that dist(M, Cm) > ?• Let B'x be the open unit
ball of X, Since (M + sBx) H Cm — 0, by the separation theorem there is
x*M G Sx* such that
s\ip{x*M(x); x eM + sBx} < mf{x*M(x)] x G CM}-
Since M is a closed subspace of X and x^ G Sx* > we have
xm\m ~° and xm(^) > ? for every x ? Cm- (*)
Put A = {arj,; M G M}. Clearly, A) is satisfied. If B = {zj^,*^, • • ¦}
is a countable subset of A, we take x G H^M,, and B) follows from the
i
second part of (*).
?
Proof of Theorem 12.41: Assume that X does not have property C.
Find A by Lemma 12.42. Then, by considering finite-dimensional subspaces

408 12. Topics in Weak Topology
of X, we can see that 0 G A by A) in Lemma 12.42. However, by B) in
Lemma 12.42, 0 g cou.vw*(B) for any B C A, B countable.
Assume now that X has property C. Let A C Bx* and / G A' . For
x* G A, put Cx* = {i E I; ?*(z) > f(x) + 1}. The sets Cx* are closed
convex and fllC^*; x* G ^4} = 0. Hence, there is a countable set B C A
with DiC**; ** ^ B) = 0- Then / € conv^B). Indeed, otherwise, by
the separation theorem, there is x G X such that g(x) > f(x) + 1 for all
g G conv™ (B). Thus z G p|{^*i x* ^ -^}> a contradiction.
?
By a "two-arrow space" we mean the compact space K — {x G [0,1]} U
{x+; x G [0,1]} ordered by the lexicographic order (i.e., x < x+ < y
whenever x < y in [0,1]). The topology of K is the order topology; that
is, a basis is given by the intervals [x+,y) and (z,x] (see, e.g., [DGZ3] or
[Fab]).
Let D be the subspace of ^[0,1] formed by all bounded real-valued
functions on [0,1] that are right continuous and have finite left limits.
Theorem 12.43 (Corson [Cor])
(i) D is isomorphic to a C(K) space, where K is the two-arrow space. D
is not weakly Lindelof but has property C.
(li) The quotient ?)/C[0,1] is isomorphic to co[0,1].
(Hi) The dual space D* is w*-separable.
Since both C[0,1] and c0[0,l] are WCG, it follows that the WCG
property and the Lindelof property are not three-space properties.
PROOF: (ii): Define <j):D/C[0}l] -» co[0,1] by </>(?):t h-> x(t) - x(t~). <j> is
well defined and ||^(?)|| = 2||x|| for all a? G D\ hence </> is an isomorphism
of D/C[Q, 1] onto co[0,1] because it has dense range.
(i): For t G [0,1], define xt G D by xt = X[t,ih wnere X[t,i] denotes
the characteristic function of the interval [t, 1]. Then S — {xt\ t G [0,1]}
is a closed discrete subset of D in its tu-topology. Note that cardE) =
2N and thus there are 22 continuous functions on S. If D were normal
in its weak topology, by Tietze's extension theorem we would obtain 22
distinct continuous functions on D in its w-topology. This is impossible for
the following reason: D*/C[0, l]1 is isomorphic to C[0,1]* and C[0, l]1 is
isomorphic to c0[0,1]*. The cardinality of C[0,1]* is 2N because C[0,1] is
separable, and the cardinality of co[0,1]* is 2N because every element of
co[0,1]* has a countable support. Thus the cardinality of D* is 2N. Hence,
the cardinality of all functions on D that are continuous in the u>-topology
of D is 2N due to Theorem 12.32. Thus, D in its w-topology is not normal
and therefore not weakly Lindelof ([Bau], [Rez]).
D has property C because it is a three-space property and both C[0,1]
and c0[0,1] have it (as WCG spaces).

12. Topics in Weak Topology 409
(iii): To see that D* is w*-separable, consider rational points in [0,1].
?
Corson Compacts
Definition 12.44
A compact space K is called a Corson compact if K is homeomorphic to a
subset S of[— l,+l]r compact in its pointwise topology such that for every
x G S we have card{7 G T; x(j) -fi 0} < No-
Every Eberlein compact is a Corson compact. In particular, the dual
unit ball of every WCG space is a Corson compact in its iu*-topology
(Theorem 11.16). We refer to [Fab] for examples of Corson compacts that
are not Eberlein compacts.
The space [0,cji] in its usual order topology is not a Corson compact
because every Corson compact is angelic (Exercise 12.55) and ui\ is not a
limit of any sequence of smaller ordinals.
Under the continuum hypothesis, there are nonseparable Corson
compacts with property CCC ([ArMe], [AMN]). Such compacts do not contain
any dense metrizable subset by the proof of Corollary 12.9. However, we
have the following result.
Theorem 12.45 (Shapirovskii [Sha])
Every Corson compact K contains a dense set formed by Gs points of K.
Proof: (Kalenda [Kail]) We first show that K contains at least one Gs
point. To this end, let K C [—1, l]r and define a partial order on K by
x < y if y = x on the support of x. Then use the Zorn lemma to find a
maximal element xq in K. By the definition of the partial order, {xq} =
{z G K\ z — xq on supp(zo)}. Since supp(z0) is countable, it follows that
the right-hand-side set is a Gs set in K. This shows that xq is a Gs point
ofK.
Note that if in general G\ D F\ D G2 D ^2 D • • •, where G; is open and
F{ is closed for all i, then f]Gi = f]F{ is a closed Gs set.
Let U be an open set in K. By the preceding remark, there is a Gs closed
set H inU. Since H is a Corson compact, it has a Gs point x\. Because H
is a Gs set in K, it follows that x\ is a Gs point of K.
?
Theorem 12.46 (Alster, Pol [AlPo], Gul'ko [Gull])
Let X be a Banach space. If (Bx* ,w*) is a Corson compact, then X is
weakly Lindelof

410 12. Topics in Weak Topology
Proof: If (Bx*, w*) is a Corson compact, then X satisfies the assumption
of Theorem 12.33 ([Fab]).
?
Lemma 12.47
Let K C [0, l]r be a compact set such that card{y G T; x(y) / 0} < No /or
ever?/ x E K. For every infinite countable subset J ofT, there is a countable
subset JofT such that J C J and Rj C K, where
RA*)W-[ 0 f0rytj.
Denote Kj = Rj(K). Then the operator defined on C(K) by Pj(f) =
f o Rj is a norm-one projection on C(K) with separable range.
Proof: (Sketch) Put J0 = J. Let Do be a countable set in K such that
Rjo(Do) is dense in Rj0(K). Put J\ — (J supp(#). Then J\ is countable,
x?D0
and we may assume that J\ D Jo- Furthermore, Rj0(K) = Rj0({x G
K\ supp(x) C Ji}).
We continue by induction and obtain an increasing sequence {J;} of
countable subsets of T such that Rjt(K) — Rj^fa E K\ supp(x) C Ji+i})
for every i. Put J = (J Jn. Note that Rj(x) — lim(Rjn(x)) for every x ? K.
Now, for every n and every x ? K, Rj^(x) G i?jn({x G iv; supp(#) C «/}).
Denote A — {x ? K] supp(x) C J}. Then, given x G if, for every n
there is xn G A such that Rjn(x) — Rjn(xn). Let y be a cluster point of
{xn} in A. Then Rjn(x) — Rjn(y) for every n. Hence Rj{x) — Rj(y) — y,
and thus iEj(x) CAcK. Therefore Rj(K) C K.
The separability of P(C(K)) follows from the fact that [0, l]1 has a base
of its pointwise topology of the cardinality card(I).
Theorem 12.48 ([Ori], [VWZ])
Let a Banach space X admit a Markushevich basis {xa]fa}. Then the
following are equivalent:
(i) X is weakly Lindelof
(ii) X has property C.
(Hi) (Bx*,w*) is a Corson compact.
If the conditions (i)-(iii) are true for a nonseparable Banach space X with
a Markushevich basis, then X* is not w*-separable.
PROOF: (i) => (ii) is immediate and (iii) => (i) is by Theorem 12.46.
(ii) => (iii): It is enough to show that card{a; f(xa) z/L 0} < Ko for
every / G Bx*- Assume by contradiction that there are / G Bx*, e > 0
and xp so that f(xp) > e for every f3 < u\. Define
Cp — span{z7; /3 < j < ui} H {x G X] f(x) > e}.

12. Topics in Weak Topology 411
Then Cp is closed and convex for every /?, and if x G f] Cp, then fa(x) =
J8<wi
0 for all a G I by the definition of a Markushevich basis. Therefore, x = 0
because {/a} is separating. This contradicts f(x) > e. Hence p| Cp = 0.
]9<wi
On the other hand, if /% < u\ for all i G N and /? < cji, /? > sup/?;, then
#0 G f]Cpr Thus, X does not have property C.
i
D
Definition 12.49
W^e wz// 5a?/ thai a Markushevich basis {xQ]fa} of X is weakly Lindelof if
{xa} U {0} is Lindelof m Us relative weak topology.
Recall that ^(r) denotes the closed subspace of ^(T) formed by all
functions that have countable support.
Theorem 12.50 ([Ori], [VWZ])
Let X be a Banach space. The following are equivalent:
@ (Bx*,w*) *"s a Corson compact.
(ii) There is a set T and a bounded linear one-to-one operator from X*
into ?%o(T) that is w*-to-pointwise continuous.
(Hi) X is weakly Lindelof and admits a Markushevich basis.
(iv) X has property C and admits a Markushevich basis.
(v) X admits a weakly Lindelof Markushevich basis.
Proof: We will sketch a proof under the assumption dens(X) = Ni.
(i) => (iii): If (?x*, w*) is Corson, then X is weakly Lindelof by the
preceding theorem. X also has a PRI ([Fab]), so it has a Markushevich
basis by the proof of Theorem 11.12.
(iii) => (iv) is trivial.
(iv) => (ii): Assume that {xa\ fa}aer is a Markushevich basis with {xa}
bounded. Define a bounded linear operator T: X* —+ ^(r) by T(f) =
(f(xQ))Q. We claim that T maps X* into ^(T). Indeed, let S be the
collection of all elements of X* that have countable support on {xa}. Note
that S is a closed subspace of X*. By Theorem 12.41, it follows that SdBx*
is u>*-closed in X*. By the Banach-Dieudonne theorem, S is w*-closed in
X*. Since S contains all fa and span{/a} is u>*-dense in X*, we get that
S = X\
(ii) => (i) is trivial.
Thus, (i)-(iv) are equivalent.
(iii) => (v): Note that if {xa;fa} is a Markushevich basis of a Banach
space X, then {^a}U{0} is weakly closed in X. (v) follows, since a closed
subspace of a Lindelof space is Lindelof.
(v) => (ii): Let {xa\ fa}aer be a weakly Lindelof Markushevich basis
of X. Given e > 0 and / G X\ let U(f,e) = {x G X; |/(x)| < e}
and, for a G I\ let Ua = {x ? X] \fa(x)\ > 0}. By the hypoth esis, the
cover {Ua,U} of {xa} U {0} has a countable subcover. Since xa ? Up if

412 12. Topics in Weak Topology
a ^ /?, it follows that all but countably many xa are in U(f,e). Thus
card{a G T; /(#a) ^ 0} < Ko- Normalize {xa} so that {a?a} C Bx and
define an operator T from X* into ^(r) by T(f) = (f(xQ))Q.
D
The proof of (iv) => (ii) also gives the following statement.
Proposition 12.51
Let {xa] fa}aer be a Markushevich basis of a Banach space X. If(Bx*, w*)
is a Corson compact, then card{a G T; f(xa) / 0} < No for all f G X*.
Corollary 12.52
Let X be a Banach space. If X* is w*-separable and (Bx*, w*) is a Corson
compact, then X is separable.
Proof: By Theorem 12.50, assume that {#a;/cjaer is a Markushevich
basis of X. Let S be a countable set that is u>*-dense in X*. Then S
separates points of X and therefore V = [J {a G T; f(xa) ^ 0}.
fes
Since card{a G T; f(xa) / 0} < Ko for all / G X* by the above
proposition, we have that T is countable, which proves the corollary.
?
Remarks
(i) Let Y be a nonseparable closed subspace of the space D from
Theorem 12.43. Then Y does not have any Markushevich basis, because its dual
is w*-separable and Y has property C.
(ii) Under the continuum hypothesis, Kunen constructed an uncountable
separable scattered compact K such that every subset of C(K) is weakly
Lindelof and every subset of C(K)* is w*-separable ([HSZ], [JiMo], [Neg]).
By Theorem 12.50, no nonseparable closed subspace Y of this C(K) admits
a Markushevich basis since Y* is inseparable (use the restriction map).
Recall that a topological space T is called pseudocompact if every
continuous real-valued function on T is bounded, and equivalently if every
continuous real-valued function on T attains its supremum over T.
Theorem 12.53 (Preiss, Simon [PrSi], Ptak [Pta], Valdivia [Vail])
Let X be a Banach space. If K C X is pseudocompact in the weak topology
of X, then K is weakly compact in X.
In the proof, the following lemma is used.
Lemma 12.54 (Preiss, Simon)
Let xq be a non-isolated point in a weakly compact subset K of a Banach
space X. Then there is a sequence {Un} of open sets in K that converges
to xq (i.e., for every neighborhood U of xq in K there is no G N such that
UnCUforn> n0).

12. Topics in Weak Topology 413
Proof: By Corollary 11.15, K is homeomorphic to a weakly compact
subset of co(r) for some set T, considered in its weak topology. We will thus
assume that K C cq(T) and x1 E [0,1] for all x E A and 7 E I\ Moreover,
we may assume that xo = 0. This is seen by replacing rby A = rx{0,l}
and defining for x E K: ?G,0) = (x7 — (xoO)-f, ?G,1) = ((#0O — #7)+-
We use again T for the index set.
We will inductively define finite subsets I\ of T (possibly empty) and
open subsets Un and Vn of A as follows.
Put 1\ = 0, Ui = Vi = A. If rt-, [/;, and V{ have been defined for
{n-l >j
x E A"; x1 < 1 for all 7 E U r* f •
Consider the two alternatives: either
A) x1 < ^ for all x E Vn, 7 E T; or
B) there exists #i E Vn and 71 E T such that (#iOl > ^-.
In case A), set Tn — 0. In case B), let F\ = {71} and suppose that
an infinite strictly increasing sequence F\ — {71} C F2 = {71,72} C •..
of finite subsets of T and a sequence #1, #2, ¦ ¦ • °f elements in Vn can be
defined in such a way that ?mG;) > ^, m E N, z — 1,..., m. This is clearly
impossible, because (xm) has cluster points in co(r). It follows that, in case
B), m E N can be found such that if x E Vn satisfies x1 > ^ for all 7 E Fm,
then z7 < ~ for all 7 ^ Fm. Define Tn = Fm.
In both cases, C/n = {x E V^; x7 > ^ for all 7 E Tn} is a nonempty
open subset of Vn such that x1 < - for all 7 ^ Tn.
We claim that {Un} converges to 0. Let U be a neighborhood of xq = 0
in A" of the form U — {x E K\ x1 < ~ for all 7 E To} for some finite set
To in T and no E N. Since {r2} are disjoint, there is mo > no such that
Tm D To = 0 for all m > rriQ. Then, given m > mo, from the definition of
Um we get x(y) < ^ < ~- < ^- for all ic E C/"m and 7 ^ Tm. It follows
that C/m C f/ if m > mo.
D
Proof of Theorem 12.53: Let K be a pseudocompact set in the weak
topology of a Banach space X and / E X*. Then sup(/) = sup (/)
K cotwIk)
and there is xq E A such that f(x0) = supK(f). Hence every / E X*
attains its supremum over conv(A), and conv(A') is thus weakly compact
by Theorem 3.55. Thus, it suffices to show that K is weakly closed.
Let xq E A \ A. By Lemma 12.54, there is a sequence {Un} of open sets
in K that converges to xq. For n E N, let xn E Un fl K. Since the weak
topology on X is completely regular (Exercise 3.54), there is / E C(K)
such that /(A \ Un) — 0, 0 < / < n on A, and /(arn) = n. Note that for
every x E A there is a neighborhood V of # in A and no E N such that
Vfl [/n = 0 for all n > no- Therefore, X^/n is well defined, continuous, and
unbounded on A, a contradiction with the pseudocompactness of A.
?

414 12. Topics in Weak Topology
We already saw in Chapter 3 that (Bx, w) is a compact space if and only
if X is reflexive, and (Bx,w) is metrizable if and only if X* is separable
(Theorem 3.31 and Proposition 3.28).
Let X* be separable, let {x*} be norm dense in Sx*, and consider on
Bx** the metric />(***,!/**) = ?2~*V*(*t*) ~ y**«)|. This metric is
compatible with the w*-topology on X**. (Bx, /?) is thus totally bounded as
a subspace of the compact metric space (Bx**, p), so if (i?x, p) is complete,
then it is compact and the space X is reflexive. However, (Bx,w) can be
metrizable by some other complete metric than p without X being reflexive.
In fact, we have the following result.
Theorem 12.55 (Godefroy [Godl])
Let X be a Banach space. If X** is separable, then (Bx,w) is a Polish
space.
Recall that a metric space is called a Polish space if it is separable and
metrizable by a complete metric (i.e., if it is homeomorphic to a separable
complete metric space). Note that the unit ball of a separable space X can
be a Polish space in its weak topology without X** being separable. An
example of this is the predual of the James tree space ([EdWh]).
Proof: Let X1 = {x*** G X***; x***(x) = 0 for all x G X}, and consider
X1 in the w*~topology from X***. Since (Bx***, w*) is a metrizable
compact, it is separable and we can find a dense sequence {yn} in (Bx±,w*)-
By Goldstine's theorem, for every n there is a sequence {#* k}k C Bx*
such that x*nk ^ yn in X***. By Exercise 2.21, X = (Xx)_l, where (.)±
denotes the annihilator in X** of a set in X***. Therefore,
Bx = (Bx**)nX = Bx**n(XL)±
= K* G Bx**; Vni***) = 0 for every n}
oo oo oo
= n n u(*"eB*-*; i*"«*)i<?}-
n = l m=l k—m
Hence Bx is a G$ set in (B^*,ti;*). Since (B^*,^*) is a compact metric
space, (Bx,w) is metrizable by a complete metric by the Mazurkiewicz
theorem ([Roy]). Finally, (Bx,w) is separable because X is a separable
Banach space.
?
Therefore, (Bx, w) is a Baire space if X** is separable. Recall that a
topological space T is a Baire space if the intersection of any countable
family of sets that are open and dense in T is dense in T (see, e.g., [AaLu]).
Concerning the Baire property for (J9x,w), we a^so nave ^ne following
result.

12. Topics in Weak Topology 415
Proposition 12.56
Let (X, || • ||) be a Banach space. If\\ ¦ \\ has the Kadec-Klee property, then
(Bx,™) is a Baire space.
Proof: Assume without loss of generality that X is infinite-dimensional.
By Exercise 3.56, Sx is a dense G& set in (Bx,w). It is easy to see that a
topological space is a Baire space if it contains a dense set that is Baire in
the induced topology. In our case, this is satisfied because on Sx the weak
and norm topologies coincide and Sx in the norm topology is a complete
metric space.
?
Since the canonical norm of l\ has the Kadec-Klee property
(Exercise 5.58), we have that Btx is a Baire space in its weak topology. The
situation in cq is different.
Proposition 12.57
BCo is not a Baire space in its weak topology.
Proof: For n G N, let Bn = {x - (xf) G BCo; \x{\ < \ for i > n}. Then
Bn is closed in BCo in its weak topology, which coincides there with the
pointwise topology. Moreover, Bn is nowhere dense in this topology. Indeed,
if x G Bn andp G N, e > 0 are given, there is x G BCo such that \xi~X{\ < e
oo
for i < p and |#max{p+i,n
}| > |, so x <? Bn. Thus, BCo = (J Bn is of the
n = l
first category in itself in its weak topology, and hence not a Baire space.
?
Like any separable space, Co can be equivalently renormed by a locally
uniformly rotund norm (Theorem 8.17). Let B\ be the unit ball of such a
norm on Co. Assume without loss of generality that J5Co C B\. By
Proposition 12.56, (Biiw) is a Baire space. However, (Bi,w) is not metrizable by
any complete metric because otherwise (BCo,w), being a closed subspace
of (i?i, w), would be metrizable by a complete metric and thus it would be
a Baire space.
Let {7V7}7Gr be an uncountable collection of infinite subsets of natural
numbers such that N7 fl Np is finite for 7 ^ /? (see, Lemma 5.16). Let
X be the closed subspace of l^ (in the sup-norm) spanned by Co and the
characteristic functions XNy of the sets N7.
Theorem 12.58 (Aharoni, Johnson, Lindenstrauss; [AhLl], [JoLl])
(i) X/cq is isometric to cq(T). X is isomorphic to C(K), where K is a
separable scattered compact such that K^> — 0.
(ii) X is Lipschitz equivalent to c$(Y).
(iii) X admits a C°°-smooth norm, has property C, and is not weakly
Lindelof.

416 12. Topics in Weak Topology
(iv) X does not contain any nonseparable closed subspace with a Marku-
shevich basis. In particular, X does not contain any subspace isomorphic
to co(T) or ip(T) for T uncountable.
(v) Bx* is w* -separable. However, there is an equivalent norm on X whose
dual ball is not w*-separable.
Since X is not weakly Lindelof, X and co(F) in their weak topologies are
not homeomorphic.
Proof: (Sketch) (i) follows form the fact that for any 71,..., jn E T and
11 n II
ai,..., an we have ^T, QjXNy. — max \a,j\. The space K is the disjoint
union of N U T U {00} topologized by letting all points of N be discrete and
the neighborhoods of 7 E T be the sets that contain 7 and sets N7\S for
finite S C Ny. 00 is then the compactification point of N U I\ It is easy to
see that X is isomorphic to Co(K), which in turn is isomorphic to C(K).
(ii): First, we prove that there is a Lipschitz selector of X/co to X\ that
is, a map if): X/co —> X such that q(ij)(x)) — x for x E X/co, where q is the
quotient map.
To obtain such a selector, we identify q(xNy) with e7 in c0(T) and define
first such a selector / on Cq~(F), the positive functions in co(r), as follows.
00
If y E CQ~(r), write y — J2 ajeiji where a\ > 0,2 > .... Put Mi = N7l
i=i
and inductively Mn — Nln \ (J Nlj. Note that XMj E X because it differs
j<n
from XNy. by an element of Cq. Let f(y) = YlajXMj- Note that q(xMj) —
<l(XNyj) = e7j and thus q(f(y)) = y.
To see that / is Lipschitz, note that f{y)n — o-i if and only if n E
N7i \GV7lU.. .UA^^). By the monotonicity of {a2}, if An = span{e7; nE
iV^}, then (f(y))n — dist(y, An) and / is thus Lipschitz. Define a Lipschitz
selector for y E c0(T) by /(y) = /(y+) - /(y~), where y = y+ - y~ is
the canonical representation of y as a difference of two disjointly supported
nonnegative terms. We have
II/M-/COII < ll/(y+)-/(*+)ll + ll/@-/(sr)ll
< 2max(l|j/+ - z+\\, \\y~ - z"||) < 2\\y - z\\.
Now define a map </?: X -+ cq 0 X/co by y?(x) — (x — <p(x), x). If (y, ?) E
Co 0 X/coj then <?>(# + y) = (y,#), so <?> is onto. Thus (p is a Lipschitz
homeomorphism of X and Co 0 X/co, which is isomorphic to Co 0 co(F),
which is in turn isomorphic to co(r).
(iii): Property C follows from Theorem 12.37. The existence of a C°°-
smooth equivalent renorming of X was proved in [DGZ3]. Pol showed that
if K is a separable compact such that if (w°) = 0 and C(K) is weakly
Lindelof, then C(K) is separable ([Poll]).
(iv): If a nonseparable closed subspace Y had a Markushevich basis, then
y* could not be ^-separable because it has property C. However, since K

12. Topics in Weak Topology 417
is separable, X* is inseparable and thus, because of the restriction, so is
y*.
(v): Since K is separable, Bx* is ^-separable by the Krein-Milman
theorem. Let || • || be an equivalent norm on X such that its dual norm
|| • ||* is LUR ([DGZ3]). Let B be the unit ball and S the unit sphere for
|| • || . Assume that there is a countable set C C B such that C = B. By
the LUR property, it follows that S C C and thus X* would be separable,
a contradiction.
?
Note that if X is Lipschitz equivalent to cq , then X is linearly isomorphic
to Co (Godefroy, Kalton, and Lancien; see, e.g., [BeLi]).
There is a Banach space X such that X* is inseparable and yet there
is no equivalent norm on X whose dual unit ball is w*-separable ([JoLl]).
Theorem 12.58 (iv) should be compared to Deville's result that every
separable infinite-dimensional Banach space with a C°°-smooth norm
contains an isomorphic copy of cq or some ^, p an even integer (see, e.g.,
[DGZ3], Chapter V).
For more information on the subject of Chapter 12, we refer for instance
to [Ziz].
Exercises
12.1 Let (T, r) be a topological space, and let p be a metric on T. We say
that (T, r) is fragmented by p if for all e > 0 and every subset A ^ 0 of T
there is a r-open set U in X such that U fl A ^ 0 and />-diam(?/ C\A)<e.
Let X be a Banach space, and let p be its canonical metric. We say that
M C X is norm-fragmented if the space (M, w) is fragmented by p.
Let K be a compact space and p be the canonical metric of the Banach
space C(K). We say that a subset M of C(K) is norm-fragmented if the
space M with pointwise topology is fragmented by p.
(i) Show that every w-compact set in a Banach space is norm-fragmented,
(ii) Show that every pointwise compact set in C(K) is norm-fragmented.
Hint: (i): Theorem 8.28. (ii): If the set C is also bounded, then it is
incompact (Theorem 12.1) in C(K) and (ii) follows from (i). For a general
set, it is enough to show that for every pointwise closed subset D of C
and every n G N, the set D fl nBc(K) is norm-fragmented. Indeed, the
fragmentability of C then follows by the Baire category theorem.
12.2 Let K be an Eberlein compact. Show that K is separable if and only
if A' is metrizable.
Hint: If K is metrizable, then C(K) is separable, so {Bc(K)*, w*) ls
a metrizable compact. Since K is homeomorphic to a subspace of

418 12. Topics in Weak Topology
(Bc(K)* > w*), K is separable. If K is separable, then, by the Krein-Milman
theorem, (Bc(k)* > w*) is separable because the extreme points of Bq(k)*
are points of if, and hence ^*-dens(C(/\)*) < No- Since C(/f) is WCG,
(^(if) is separable by Proposition 11.3. Then (BC(K^ w*) and thus also K
are metrizable.
12.3 Illustrate Theorem 12.4 on (Bi2(r),w).
Hint: Sx is metrizable by the norm metric since on Se2(r)> the norm and
weak topologies coincide.
12.4 Show that G$ points of (i^2(r)> w), T uncountable, are exactly S^r)-
Hint: Follows similarly as the proof of Sx — Bx for infinite-dimensional
spaces (Exercise 3.8).
12.5 Let K = [—1, l]r in its pointwise topology, T uncountable. Does K
contain a G$ point?
Hint: No, use the definition of the topology in K.
12.6 Does (BCo(r),w) contain a G§ point?
Hint: No.
12.7 Show that every compact metric space K is a uniform Eberlein
compact.
Hint: C(K) is separable; if {xn} C Sc(K) is dense, the map T: C(K)* —> ?2
defined by T(f) = (f(xi)/2l) is one-to-one and u^-ru continuous. K is
homeomorphic to a subset of Bc(k)* m its it;*-topology.
12.8 Let if be a uniform Eberlein compact, and cp a continuous map from
K onto a compact L. Show that L is a uniform Eberlein compact.
Hint: C(L) is isomorphic to a closed subspace of C(K). Hence C(L) has a
UG-smooth norm.
12.9 Show that every WCG Banach space X is generated by a set that is
a uniform Eberlein compact in the weak topology.
Hint: Let T:X* —> co(T) be a bounded one-to-one w*-^-continuous
operator onto a dense set in co(T). Then X is generated by T^B^r))- The
formal identity map of ?\(T) into ?2(r) shows that B^r) is a uniform
Eberlein compact in its iu*-topology.
12.10 Let M C X be a bounded closed convex set that satisfies the
following: If fnign e Sx* satisfy ||/n + gn\\ -* 2, then sup (fn - gn)(x) -> 0.
Show that M is weakly compact.

12. Topics in Weak Topology 419
Hint: Use the sets 5n>p(n) from the proof of Lemma 12.19 and note that
M C f! Sn,P(n) CX H f]Sn)P{n)w C X. Thus F CI, showing that M
n n
is weakly compact.
12.11 Following the hint, show that there exists a scattered Eberlein
compact K that is not a uniform Eberlein compact.
CO
Hint: Let Y — f|{l,...,n}.A subset A of T is said to be admissible if it
n = l
satisfies the following condition: There exists a natural number n — n(A)
so that for any x,y ? A with x ^ y we have x(j) = y(j) for j < n — 1 and
z(rc) ^ 2/(™).
Consider if = {xa\ A C T admissible} with the topology of pointwise
convergence. K is an Eberlein compact: We can naturally embed K into
co(T) with the w-topology. Hence, we need only show that K is u;*-closed
in ^oo(r). This is true because if A C T is not admissible, then there exist
ji E A, i — 1,2,3, such that the set {71,72,73} is not admissible. Hence,
G = {/ E 4o(r); f(ji) > \yi — 1,2,3} is an open neighborhood of xa
that does not intersect K.
K is not a uniform Eberlein compact: Assume the contrary. By
Theorem 1.8 in [ArFa], where we put X — co(r), x1 = e7, e = |, we
obtain a decomposition {1^}^! of T and numbers {k(n)}^L1 so that
card(A D Tn) < &(n) for any admissible set A and n E N. Because T is a
00
complete metric space and r = (J Tn, by the Baire category theorem there
n = l
00
exists no E N and a basic open subset V = (ni,..., n^) x Yl {!> • • • > n)
n=ib + l
of T such that Tno n V is dense in V. Hence Tno contains an arbitrary large
admissible set, which is a contradiction.
K is scattered: Let A = {71,..., jn} C T be an admissible set. Consider
G = {x E K] x(ji) = 1, i = 1,..., n} C K C c0(r). If 5 C T is such that
Xb G G, then card(I?) > n and {7;; i = 1,..., n} C B. Hence, if n(yl) = n,
then the set G defined above contains no other characteristic function of
an admissible subset of T. Consequently, A is an isolated point of K. By
induction,
#(») \ #(n+1) = {xa] A admissible, card(A) > 1, n(A) - card(A) = n};
therefore #(w°) \ X(^+i) = {Xa] a admissible, card(yl) = 1}, tf<>0+1) =
{x0},and/^°+2) = 0.
12.12 Prove that every scattered compact is sequentially compact.
Hint: Theorem 12.29 and Theorem 10.10, which is also valid for Asplund
spaces.
12.13 Let K be an infinite scattered compact. Show that C(K) is
isomorphic to its hyperplanes.

420 12. Topics in Weak Topology
Hint: Theorem 12.30.
12.14 Let K be a scattered compact. Show that every weakly compact
operator from C(K) into a Banach space X is a compact operator.
Hint: Theorem 11.28, Theorem 12.28, and Schur's property of ?i(T).
12.15 Let T be an infinite set. Let K = T U {oo} be an Alexandrov com-
pactification of the discrete space V. Show that K is a scattered Eberlein
compact and C(K) is isomorphic to co(r).
Note that K is a uniform Eberlein compact by Theorem 12.17.
Hint: K is Eberlein since co(T) is weakly compactly generated
(Theorem 12.12). From the definition of the scattered compact, it follows that K
is scattered.
If 7 G T, then H = {/ e C(K)\ f(j) = 0} is a hyperplane in C(K)
which is isomorphic to C(K). This is seen similarly as in cq\ use the map
(a?i, #2,...) h-> @, ari,...). Since all hyperplanes of a given Banach space are
mutually isomorphic (Exercise 2.7), C(K) is isomorphic to the hyperplane
{/ G C(K)\ /(oo) = 0}, which is in turn isomorphic to Co(r).
12.16 Let K be a compact metric space. Show that K is countable if and
only if C(K)* is separable.
Hint: First, note that C(K) is separable since K is a compact metric space
(Lemma 3.23). If K is countable, then K is scattered by Lemma 12.25.
Therefore, C(K)* is separable by Theorem 12.29. If C(K)* is separable,
then K is scattered by Theorem 12.29. Since K is moreover metrizable, K
is countable by Lemma 12.25.
12.17 Let K\,K2 be compact spaces. Assume that C(K\) and C(K2) are
isomorphic.
(i) If K\ is an Eberlein compact, is K^ necessarily an Eberlein compact?
(ii) If K\ is scattered, is Ki necessarily scattered?
(iii) If K\ is countable, is Ki necessarily countable?
Hint: (i): Yes, C{K.2) is weakly compactly generated.
(ii): Yes, see Theorem 12.29. . .
(iii) Yes. K\ is metrizable by Lemma 12.25. Therefore, C{K\) is separable
by Lemma 3.23. Thus C^K^) is separable. Since K\ is scattered, C(K\) is an
Asplund space by Theorem 12.29. Thus K^ is scattered by Theorem 12.29.
Since C{K-2) is separable, Ki is metrizable by Lemma 3.23. Thus Ki is
metrizable and scattered, and hence countable by Lemma 12.25.
12.18 For a scattered compact K, let a(K) be the smallest ordinal such
that R(a(K^ = 0. The Bessaga-Pelczyriski theorem [BePl] asserts that if
Ki,K2 are infinite countable compact spaces such that a(Ki) < a(K2),
then C(K\) is isomorphic to C(K2) if and only if there is n G N such that
a(^2) < {oc{Ki))n. This theorem compares with the result by Milyutin

12. Topics in Weak Topology 421
(see, e.g., [Woj]) that if K is an uncountable compact metric space, then
C(K) is isomorphic to C[0,1].
Use the quoted Bessaga-Pelczyriski theorem to derive that for a compact
space K, C(K) is isomorphic to Co if and only if K^0^ = 0.
Hint: cq is isomorphic to C(ivo), where i^o is an Alexandrov compact-
ification of N, so C(K) is isomorphic to C(Ko) and, by the previous
exercise, A" is countable; it is also infinite. Since K is compact, we have
a(K) > 2 = a(/\o). Thus we can use the Bessaga-Pelczynski theorem. To
finish the proof, it is enough to observe that, by compactness, for a compact
set L we have L^0) = 0 if and only L^ — 0 for some positive integer n.
12.19 Let K be an infinite metrizable compact space. Prove that:
(i) C{K)* is separable if and only if C(K) does not contain a subspace
isomorphic to t\.
(ii) unexposed points of Bq{k)* form a James boundary of C(K).
Recall that / G Bx* is a unexposed point of Bx* if there is x G Sx
such that f(x) = 1 and z is a point of Gateaux smoothness of the canonical
sup-norm of C(K).
Hint: (i): If C(K)* is not separable, K is not scattered (Theorem 12.29)
and thus C(K) contains a subspace isomorphic to C[0,1]. The space C[0,1]
contains an isomorphic copy of t\ (Theorem 5.17).
(ii): Given / G Sc(k), find ^o G K is such that f(to) — 1. Define a
function <p on K by (p(t) = 1 — dist(tf,^o). Then <p is a point of Gateaux
differentiability of the sup-norm on C(K) by the Smulian lemma. Hence t0
is a w*-exposed point of Bc(k)* • Thus, unexposed points of Bx* form a
James boundary of C(K).
12.20 Find an example of a compact set K such that C(K) is nonseparable
but does not contain cq(T) for any T uncountable.
Hint: /?N, because t^ has a ^-separable dual unlike co(r), T uncountable.
12.21 Show that if C(K) is reflexive, then K is finite.
Hint: c0 in C(K)?
12.22 Show that a continuous image of a metrizable compact is metrizable
by using C(K) spaces, subspaces, and their separability.
Hint: If L — <p{K), then C(L) is a subspace of C(K) and C(K) is separable.
12.23 Show that /?N does not contain any infinite metrizable subset.
Hint: Otherwise, C(/?N) would have a nontrivial convergent sequence and,
by Tietze's theorem, ^oo = C(/?N) would factor to Co, which is not the case
(Exercise 6.31).
12.24 Let K be a metrizable compact. Show that C(K) admits an
equivalent C^-smooth norm if it admits a Cfl-smooth bump.

422 12. Topics in Weak Topology
For non-metrizable spaces, this result no longer holds in general (Haydon;
see, e.g., [DGZ3]). However, it is not known whether a separable Banach
space X admits an equivalent G°°-smooth norm if it admits a G°°-smooth
bump.
Hint: C(K)* is then separable, so K is scattered. Since K is metrizable,
this means that K is countable. Then use Theorem 10.46.
12.25 A collection T of subsets of N is hereditary if G C F, F G T implies
GeT.
Prove the following Ptak combinatorial lemma ([BHO], [Pta]).
Let 7 be a hereditary collection of finite subsets of N and let 6 > 0 be
n
given. Assume that for all ai,..., an > 0 with ]T ai = 1 there exists F E. T
i=l
such that ^2 a{ > 6. Then there is an infinite subsequence M of N such
that F G T for all finite F C M.
Hint: Define a norm on coo by | Ylaiei\ ~ sup\ \\Y1 ai\\ ] F E !F>. Let X
be the completion of coo in this norm. Then {e2} is a Schauder basis of X
that is equivalent to the canonical basis oi?i. Note that T can be identified
with a closed subspace of X* by using the mapping <?(]Pa;e;) = J2 ai- ^n
i?F
this way, T is a 1-norming set in X*.
Denote by K the iu*-closure of T in X*. t\ is isomorphic to X, which
is in turn isometric to a subspace of C(K). Thus K is not countable, for
otherwise C(K) would be an Asplund space, which is not true for ?\. Since
K can be identified with the closure of T in 2N in its pointwise topology,
K contains an infinite sequence M. Because T is hereditary, every finite
subset of M is in T.
12.26 Using Ptak's combinatorial lemma, prove the following variant of
Mazur's theorem.
Let K be a compact space and /, /i, /2,... G C(K) be such that {/&}
is bounded in C(K) and /n —* / pointwise. Then for every e > 0, there
exist ni,..., rijk G N and Ai,..., Afc G R such that 0 < A; < 1 for every i,
II k II
J2Xi = 1, and /- ]T Kfnt\\ <e.
II i-l lloo
Hint: Assume that / = 0 and ||/n||oo < 1 for every n. Given e > 0, for each
x G K put Gx = {n G N; |/n(*)| > e/2}. Note that Gx is finite for all
x G if. Set F* = {G; G C Gx}, and let ^* = {i^; x G A'}. Assume that
T satisfies the conclusion in Ptak's combinatorial lemma. Then there is an
increasing sequence {n2} of natural numbers such that {ni,..., n^} C FXk
for each k. Let xq be an accumulation point of {xk}- Fix k G N. Since
^fc G i7^ for all z > k, \fnk(^i)\ > ?/2 for all i > A:. By the continuity of
fnk, we thus have \fnk(xo)\ > s/2. Hence n^ G -F^o f°r a^ & G N. This
shows that FXo is infinite, a contradiction.

12. Topics in Weak Topology 423
Thus, by Ptak's lemma there exist ai,..., a* with a; > 0 and J2ai = 1>
such that ^ a2- < e/2 for all z G A". Let x E K. Then
tGFx
= | Yl aifi(X)+ Yl °>ifi(X)
"* = 1 «6{l,...,A:}nFx 2€{l,...fc}\Fx
*6{i,...,fc}nFx *e{i,...,fc}\Fx
12.27 We denote [M]2 = {(mi,m2); mi < r«2 G M} for a subsequence
M of N. Prove the following version of the Ramsey theorem.
Let A C [N]2- Then there is a subsequence M of N such that either
[Af]2 C A or [M]2C[N]2\ A
Hint: For simplicity, denote A\ — A, A2 = [N]2 \ A. Choose any mi G N.
Then there is k G {1,2} such that {m > mi; (mi,m) G .A*} is infinite.
Hence there is a subsequence M\ of N such that the set {(mi, m); m G Mi}
is a subset of Afc for the appropriate k.
Choose any m2 G Mi such that m2 > mi. As before, find a subsequence
M2 of Mi such that {(m2,m); m G M2} is a subset of Ak for some k
(which may be different from the one for Mi). Continuing in this manner,
we obtain mt-, M{.
Let I — {z; (rrii,m) G A\ for all m G M2}. If / is infinite, set M =
{rrii}i€i and we have (a, 6) G -A for all a < 6 G M. In the other case,
N \ I is infinite, so we set M = {mt-},-^/ and have (a, 6) G [N]2 \ A for all
a< b G M.
12.28 (Root lemma) Let *4 be an uncountable family of finite sets. Show
that there is an uncountable subfamily B of A and a finite (possibly empty)
set S such that A D B = S for every pair of distinct elements A, 5 of 6.
The family B is called a A-system and S is called a root of #.
Hint: Assume that card(*4) = Ki and that .4 is formed by finite sets in
[0,u;i]. For n G N, set ^n = {A G .4; card(A) = n}. There is n such that
card(^4„) — Ni. Fix this n. If A ? An, write A — {A(l),.. .,A(n)} with
A(l) < AB) < ... < A(n).
For every a < u>i, the set {A G .4n; -A C [0,a]} is countable. Hence
sup( (J A] = wi. Let p G {l,...,n} be the least integer such that
sup{A(p); A G An} = wi. Put a0 = sup{A(p-l)+l; A G An} (if p = 1, we
put ao = 0). By transfinite induction on \i < u>i, choose A^ G .4n such that
A^p) > max{a0,sup{Ai/(n); v < //}} and set B\ = {A^; // G [0,u>i)}.
Then card(Si) — Ni and A C\ B C [0,a0] whenever A, 5 are distinct
elements of B\. Since the family of all finite sets in [0, c*o] is countable, there
is an uncountable family B C B\ and a finite set S C [0,ao] such that
A fl [0, c*o] = 5 for every A? B. Clearly, 6 is a A-system with root 5.

424 12. Topics in Weak Topology
12.29 Let X be a Banach space nonseparable in its weak topology. Does
X contain a dense metrizable subset?
Hint: No, X would not have property CCC.
12.30 Show that if a Banach space X admits a Lipschitz, Gateaux
differentiable bump, then ^ is not a quotient of X.
Hint: Otherwise, /?N is in X* and has no nontrivial convergent subsequence,
a contradiction with Theorem 10.10.
12.31 ([Pol2]) Let //bea Radon measure on a compact space K such that
C(K) has property C. Show that \i has separable support.
Hint: Assume that supp(/x) — K. We will show that K is separable.
Fix i G N. For x eK, put
Cx = {/ G C{K)- j fdli>\ and f(x) = o}.
K
If z G n CXi then / = 0 identically on K and thus f f dfi = 0, a contra-
x?K K
diction. Hence f] Cx = 0. Since Cx are all closed and convex, and C(K)
xeK
has property C, f] Cx — 0 for some countable set Ai C K. We claim
xEAi
oo
that K = |J Aj. Assume that there is / G C(Ar) such that / = 0 on |J Ai
»=i
and J/d/i > 0. Find i G N such that f fdfi > i. Then / G f| ^i a
K k l x?Ai
contradiction.
12.32 Show that dens(^) = 2C, and hence dens(^) = T > card^) = c.
Hint: The cardinality of/?N.
12.33 Let X be a separable Banach space and assume that the dual norm
of X* is Gateaux differentiable. Show that every element of X** is a first
Baire class function when considered as a function on (Bx*, w*).
Hint: Let F G Sx** attain its norm at / G Sx* • Let xn G Sx be such
that f(xn) —> 1. Then xn as functions on (Bx+,w*) are continuous and,
by Smulian's lemma, xn —> F pointwise on #x* • Thus, F is a first Baire
class function on (Bx*, w*).
By the Bishop-Phelps theorem, every G G Sx** is a uniform limit (on
Bx*) of elements of Sx** that attain their norms.
For further use of Baire methods, we refer to [Nat].
12.34 Let T be a topological space. We define the weight of T, w(T), as
the minimal cardinal K such that there is a basis T of topology of T with
card(^) < N.
Show that if K is a compact space, then w(iv) = dens(C(/\)).

12. Topics in Weak Topology 425
Hint: dens(C(K)) < w(K): follow the proof of Lemma 3.23.
vr(K) < dens(C(iv)): Assume that T is dense in C(K) and card(^7) =
dens(C(A")). The topology on K of pointwise convergence on elements of
T has a basis of cardinality card(.F).
12.35 Let X be a separable Banach space, and assume that F is a closed
subspace of X* that is w*-dense in X*. Is F w*-sequentially dense in X*?
Is it true that every element of X* is a w*-limit of a iu*-bounded net in Fl
Hint: No. Let Gn = F D nBx* ™ . Then, assuming \JGn = X*, by the
Baire category theorem and the symmetry and convexity of Gn, at least
one Gn contains a ball 6Bx* for some 6 > 0. Then F is a norming subset.
Each separable space such that dim(X**/X) is > No contains a w*-dense
non-norming closed subspace F C X* ([DaLi]).
12.36 Assume that X is a separable Banach space and {fa} C X* is a net
that converges to 0 in the topology of uniform convergence on sequences
{x{} in X that converge to zero. Is {/<*} necessarily bounded?
Hint: No. Take a iu*-dense non-norming closed subspace F in X* (see
the previous exercise). Then the closure of F in the topology of uniform
convergence on sequences converging to zero is X* ([DuSc]). Take a point
/ in X* that is not in Gn = F fl nBx* for any n (see above). Then /
can be reached by a net {fa} converging to / in the topology of uniform
convergence on converging to zero sequences. This {/<*} is not bounded,
because / is not in any Gn.
12.37 Let X be a separable space. Is X* w*-sequentially separable; that
is, does there exist a countable C C X* such that each element of X* is a
w*-limit of a sequence in C?
Hint: Yes. nBx* is metrizable, w*-separable, and X* = [JnBx*-
12.38 Show that P^ is not w*-sequentially separable.
Note that, by Goldstine's theorem, it is w*-separable.
Hint: Use the Grothendieck property.
12.39 Show that Do/co)* = cfr C t^ is not w*-separable.
Hint: co(c) C Ax>/co (Exercise 5.52).
12.40 Show that there is an equivalent norm on t^ such that its dual
ball is not w*-separable, although the standard unit ball of t^ — Cq* is
if*-separable by Goldstine's theorem.
Note that there is a Banach space X such that its dual is w*-separable
and the dual ball of no equivalent norm on X is w*-separable ([JoLl]).
Note also that the dual space C(K)*, where K is Kunen's compact, has
the property that every subset of C(K)* is w*-separable.
Hint: Extend on ?qo the norm in Theorem 12.58.

426 12. Topics in Weak Topology
12.41 Let X be a separable Banach space. Is every subspace of X*
inseparable?
Hint: Yes. Let 7bea subspace of X*. (nBx*, w*) is a separable metrizable
space, so An = Y f) nBx* is a separable space in the w*-topology. Thus,
Y = [JAn is a inseparable space.
12.42 Let fn E X* be such that ||/n|| = ^ for each n. Write each /n as
w*-lim(/?), where ||/n — f%\\ = n (Josefson-Nissenzweig). Show that the
origin of X* is not a tu*-limit of any sequence in {f^}n,k-
Hint: The Banach-Steinhaus theorem.
12.43 Prove that C[0,u;i]* is not w*-separable.
Hint: co[0,u;i] can be isomorphically embedded into C[0,o;i], and co[0,cji]*
is not ^*-separable.
12.44 (i) Show that ?i(T) does not have property C if T is uncountable,
(ii) Show that l^ does not have property C.
Hint: (i): The space ^i(r) does not have property C because it has a Marku-
shevich basis (the standard basis), and its dual ball in the w*-topology is
not Corson (not even angelic; use Goldstine's theorem and the fact that
each element of Cq(c) is countably supported).
(ii): li(c) C C[0,1]* (Exercise 3.85) and hence ?i(c) C 4o-
12.45 Let / E C[0,u>i]. Show that there is a < loi such that / is constant
on [a,ui].
Hint: For n E N, let an < o;i be such that |/(/?) - /(u>i)| < ? for all
/? > an. Consider sup(an).
12.46 Prove that the Stone-Cech compactification of the ordinal segment
[0, a) is homeomorphic to its one-point compactification.
Hint: Continuous functions are eventually constant.
12.47 For uq < a < u>i, define the projections in C[0,u;i] by
p{fm _//(/?) for/?<a,
fcKTAP) ~ | /(a) for /? > a.
Show that {Pa} is a norm-one projectional resolution of the identity on
C[0,Ul].
12.48 For a < k>i, define a projection Pa ofCo[0,u;i] by Pa(x)(f3) — x(C)
for /? < a + 1 and Pa(x)(/3) = 0if/?>a+l. Show that the projections
Pa together with the identity operator satisfy all the properties needed to
form a PRI but one, namely the continuity of the maps a »-> Pa(x) on
the ordinal segment (use the characteristic functions of [0,o;]). However,
(J Pa(Co[0)wi]) = Co[0Iwi]and U ^(CofO,^]*) = Co[0jWl]*.
a<u>i or<a;i

12. Topics in Weak Topology 427
Hint: If x ? Co[0,a;i], then x(a) = 0 for all a > ao. By this argument, the
first part in the statement follows. To see the latter part, let / ? Sc^o^i]*
be such that f(x) — 1 for some x ? Sc0[o,wi]- Let x(/3) = 0 for all C > f30.
If he Co[0,a;i], \\h\\ < f, then \\x ± h\\ < 1 and thus \f(x ± h)\ < 1. By a
convexity argument, /(x ± A) = 1, which means that /(/i) = 0. From this
we get P* (/) = /.
12.49 (Semadeni) Show that C[0,cji] 0 C[0,cji] is not isomorphic to
C[0,wi].
Hint: Show that the codimension of C[0,u>i] in the space of all sequentially
w*-continuous functions on C[0,u;i]* is 1.
12.50 Prove that C[0,tJi] does not have an unconditional basis.
Hint: Such a basis would be shrinking since the space is Asplund. Thus the
space would be WCG, a contradiction.
12.51 (i) Is ^oo isomorphic to a subspace of C[0,o;i]?
(ii) Is C[0,u;i] isomorphic to a subspace of ?oo?
Hint: (i) No, the Asplund property is hereditary ([Phe2]).
(ii) C[0,a;i]* is not u>*-separable.
12.52 A bounded linear operator T from a subspace X of a Banach space
C(K) into c0(K) is called a Talagrand operator ([Tal3], [Hay2]) if for every
# G Sx there is k ? K such that \x(k)\ — 1 and T(x)(k) ^ 0. Construct a
Talagrand operator on Co[0,o;i].
Hint: For x ? Co[0,u/i], put
v /v y [ 0 if a — ui.
Then T(x) ? co[0,cji] because x ? C[0,ui]. Let \\x\\ = 1. Choose a < w\
maximal such that \x(a)\ — 1. Then check that T(x)(a) ^ 0.
12.53 Assume that a subspace X of C(K) admits a Talagrand operator T
into co(K). Show that X admits an equivalent norm that locally depends
on finitely many coordinates.
Hint: |*I = sup{|;c(A0| + |T(s)(*)|; k ? #}.
12.54 Show that C[0,o;i] admits a norm that locally depends on finitely
many coordinates.
Hint: Co[0,u;i] is isomorphic to C[0,o;i].
12.55 Show that every Corson compact K is an angelic space.
Hint: Let K C [0, l]r be a compact space such that, for every / ? K, the
set {// ? T; /(//) / 0} is countable. Let H C K and h ? #. Let {^?} be
the support of h. Find hi ? if such that |(/i — ^i)(/^i)| < 1. Let {fi}} be the

428 12. Topics in Weak Topology
support of hi. Find/12 G H such that \(h - h2)(fi))\ < |, i = 0,1, j = 1,2,
oo
etc. The sequence hi converges to h at every point of (J supp(//). Outside
this set, h and hj are zero.
12.56 Show that every separable Corson compact is metrizable. Thus, for
Corson compacts, metrizability is equivalent to separability.
Hint: Let Kc[0, l]r be a compact such that, for every k E K, the support
of k on T is countable. The set [0, l]r is equipped with its natural product
topology. Let {kn} be dense in K. Let Ti be the union of all supports of
{kn}. Use the fact that [0, l]Pl is metrizable in its product topology.
12.57 Assume that X is a separable Banach space such that Bx** in its
w*-topology is a Corson compact. Show that X* is separable.
Hint: Since X is separable, Bx** is if*-separable by Goldstine's theorem.
Thus, Bx** in its uAtopology is metrizable by the previous exercise. Hence
X* is separable by Proposition 3.24.
12.58 Show that the ordinal segment [0,u;i] in its standard topology is a
scattered compact which is not a Corson compact.
Hint: The point uj\ is in the closure of the set of smaller ordinals, but it is
not a limit of any sequence of them. Thus the space is not angelic. Every
Corson compact is angelic (Exercise 12.55).
By the definition, it follows that [0,u;i] is a scattered compact.
12.59 Show that the space C[0,u>i] is not weakly compactly generated.
Hint: By the previous exercise, [0,cji] is not an angelic space and thus not
an Eberlein compact; use Theorem 12.12.
12.60 Verify that the alternative definitions of pseudocompact (before
Theorem 12.53) are equivalent.
Hint: f^j-
12.61 Show that the convexity assumption in Theorem 3.55 cannot be
dropped.
This shows that any characterization of general weakly compact sets via
attainment of maxima (for instance, Theorem 12.53) must consider more
functions than just elements of X*.
Hint: Consider Bt2 \ \Bi2.
12.62 Let K be the Kunen compact as in the Remarks after
Corollary 12.52.
(i) Show that C(K) admits no locally uniformly rotund norm.
(ii) Show that C(K) admits no Frechet differentiable norm.
(iii) Show that every closed subspace of C(K) is an intersection of a
countable family of hyperplanes.

12. Topics in Weak Topology 429
It is not known whether this C(K) admits a Frechet differentiate bump
or a Gateaux differentiable norm.
Hint: (i): If a norm on C(K) were LUR, then its unit sphere would be norm
Lindelof, which it cannot be because it is not separable.
(ii): In such a norm, the set S of x* G Sx* that attain their norm is a
^-separable set and on S the norm and w*-topologies coincide. Since S is
dense in Sx* by the Bishop-Phelps theorem, Sx* would then be separable,
a contradiction.
(iii): If Y C C(K), then Y1 is w*-separable.
12.63 Let K be a uniform Eberlein compact, and let ^ be a continuous
map from K onto L. Show that L is a uniform Eberlein compact.
Hint: C(L) is a subspace of C(K). Use Theorem 12.18.
12.64 Recall that a compact space K is called a Valdivia compact if K is
homeomorphic to a set Kq in some [0, l]r in its pointwise topology such
that the set of all points in Kq that have countable support is dense in Kq.
Note that every Corson compact is a Valdivia compact.
Prove that [0,u>i] and [0, l]r for every T is a Valdivia compact. For more
information on Valdivia compacts, we refer for instance to [Val2], [AMN],
[DeGo], [Kal2], [Kail], and Chapter VI of [DGZ3].
Hint: Consider characteristic functions of intervals (a,u;i] and with every
ordinal associate evaluations on those functions. For the second example,
observe that finitely supported vectors are dense in [0, l]r.

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Index
annihilator YL and Yj_, 40, 55, 58,
93, 148, 149
Banach limit, 62
basis
algebraic, 34, 191
Auerbach, 139, 164
bimonotone, 191
block, 172
boundedly complete, 166-168, 182,
192
constant, 163, 169, 182
equivalent, 169-171
Ham el (see basis: algebraic)
Markushevich, 188-190, 382,
410-412
shrinking, 188, 197, 369, 370
weakly compact, 364
weakly Lindelof, 411
monotone, 163, 191, 192
normalized, 163
orthonormal, 18, 20, 222
Schauder, 161, 163, 165, 303, 307
seminormalized, 303
shrinking, 166-168, 184, 192, 260
summing, 165, 181
unconditional, 180, 181, 196, 197
bump. See function
(*x, ||-II), 2, 14
(Bx,w), 73, 75, 414, 415
(Bx,w*), 71-73, 319, 365, 395,
409-412
cardinality card(^l), 23, 403
*
closure M , M , M, 64
compact
Corson, 409-412, 427, 428
countable, 345, 399, 420
Eberlein, 365, 367, 388, 390-393,
409, 417, 419, 420
scattered, 398-401, 419, 420
uniform Eberlein, 394, 395, 418,
419
complement, 137, 138, 147-149
algebraic, 137, 147
orthogonal FL, 17, 18, 138
quasicomplement, 377
constant
basis, 163, 169
unconditional basis, 182
conv(M), 2, 22, 85, 92, 104
convergence
in norm —», 65
pointwise, 66, 68, 86

446 Index
convergence (continued)
weak -^, 65, 86, 87, 105
weak star —> , 65, 87, 105
coordinates, 161
decomposition method, 140, 174
density dens(X), 358, 359, 381, 382,
397, 424
derivative
Cantor, 398
directional, 241
Frechet, 241, 307
Gateaux, 241
Dirac measure, 48, 78
distance, Banach-Mazur, 283
distributions, 116, 133
domain Dom(/), 317
dual. See operator; space
embedding, canonical x, 63
e-net, 30
epigraph Epi(/), 317
equality
parallelogram, 17, 29
Parseval, 20
polarization, 17
Ext(C), 76, 78, 80, 123
form
bilinear, 158
n-linear, 313
quadratic, 159
symmetric, 313
Fourier coefficient/series, 19
function
bump, 314, 315, 317-319, 321, 340,
342, 345, 346, 424
C1 -smooth, 241
C°°-smooth, 314
conjugate, 321, 347
convex, 242, 247, 253, 263, 265, 320
depends on finitely many
coordinates, 343
i^-smooth, 314
Frechet differentiable, 241, 242,
251, 253, 254, 263, 320
Gateaux differentiate, 241, 247,
263
inf convolution, 265
proper, 317
Tfc-smooth, 340
UF, 289
UG, 289
w*-differentiate, 333, 334
functional
biorthogonal, 139, 164, 192
coordinate, 164, 171, 193, 197
Dirac, 48, 78
from X**, 69, 127
linear, 11, 37, 38
Minkowski, 42, 54
sublinear, 37
supporting, 40, 242
halfspace, 64
homeomorphism, 25
Lipschitz, 331, 334, 349
uniform, 133, 352, 354
hyperplane, 53, 76, 153
inequality
Bessel, 20
Cauchy-Schwarz, 3, 16
H61der, 3
Khintchine, 178
Minkowski, 4
Simons, 80
triangle, 1
inner product, 16
Int(ilf), 31
James boundary, 79, 343, 344, 355
kernel Ker(T), 11, 58, 93, 208, 217
Kronecker 8ij, 139
Lipschitz. See homeomorphism; mar
local base, 107
map
affine, 102
contraction, 227, 237
differentiation Da, 113, 116
fixed point, 227
homeomorphism, 25
identity 7x, 11
Lipschitz, 10

Index 447
map (continued)
nonexpansive, 227, 237
open, 49, 50
retract, 239
metric, translation invariant, 108
Minkowski, 42, 54
modulus
of convexity, 285, 289
of rotundity, 285, 289
of smoothness, 288, 289
net, 65
Cauchy, 109
norm, 1
attaining, 40, 83, 102, 260, 262
^-smooth, 241, 244
C^-smooth, 314, 344, 345, 421
dual, 96, 245, 246, 249, 280
equivalent, 11, 12, 25, 55
Frechet differentiate, 242-244,
249, 250, 253, 254, 259, 260,
267, 271, 328, 367, 369
Gateaux differentiate, 242, 243,
246, 247, 250, 264, 275, 276,
367, 382
LUR (locally uniformly rotund),
248-250, 253, 271, 277, 280,
281
operator (see operator)
rotund, 246, 277, 281, 305
strictly convex, 246, 277, 281, 305
UC (uniformly convex), 285, 287,
290, 291, 293, 294, 301, 303,
305, 306, 308
UF (uniformly Frechet), 288-291,
293, 294, 304, 306, 307
UG (uniformly Gateaux), 289, 395,
397
uniformly smooth, 288, 290, 291,
293, 294, 304, 306, 307
UR (uniformly rotund), 285, 287,
290, 291, 293, 294, 301, 303,
305, 306, 308
URED, 307, 308
WUR, 397
operator
absolutely summing, 372-375
adjoint T*, 217
bounded, 10, 11, 90
compact, 175, 194, 203, 207-209,
215, 216, 222, 223, 226, 230,
232, 372
completely continuous, 206, 232,
375
Daugavet, 310
dual T*, 51, 58, 60, 90, 207, 212,
371
eigenspace, 214, 216
eigenvalue, 214-216, 222
eigenvector, 214, 221, 222
finite-rank, 152, 203
Fredholm, 209
Hilbert-Schmidt, 236
invertible, 50, 210, 211, 213
isometry, 11, 25, 60, 90, 224
isomorphism, 11, 50, 60, 90, 349
isomorphism into, 11, 25, 59
norm, 11, 51, 218
normal, 224, 226
on ?p, 59, 90, 98, 157, 175, 195,
196, 232, 274
on co, 157, 175, 176, 195, 196, 233,
274, 364, 365, 384
onto, 50, 59
resolvent p(T), R(\), 211, 223
self-adjoint, 217-223
spectral decomposition, 223, 226
spectral radius r(T), 213
spectrum
approximate, 310
a(T), 211, 212, 216, 217, 219,
220, 222, 234
strictly singular, 175, 194, 195, 235
unitary, 224
weakly compact, 370-372, 383
partition of unity, 328
point
cluster, 65
diametral, 274
exposed, 255, 276
extreme, 76, 98, 121, 276
strongly exposed, 255, 256, 262,
277
w"-strongly exposed, 278
polar A0 and A0, 118, 133
polarization identity, 17

448 Index
polynomial, 313
PRI, 359, 382, 383
projection, 137, 147
canonical Pn, 161, 163, 165, 260
Dixmier's, 148
orthogonal, 225
property
approximation, 192, 205
Banach-Saks, 89
C, 405-407, 410, 411
CCC, 391, 392, 401
drop, 272
Dunford-Pettis, 375-377
Grothendieck, 195, 383
Heine-Borel, 114
Kadec-Klee, 280, 415
w*-Kadec-Klee, 280
lifting, 141
Mazur, 383
Schur, 146, 156
three-space, 24, 97, 405, 408
Rademacher functions rn, 177
seminorm, 37, 54
sequence
basic, 169, 170
equivalent, 169-171
unconditional, 181
block basic, 172, 173
bounded, 6
Cauchy, 6
weakly Cauchy, 145
series
absolutely convergent, 8, 27, 373
unconditionally convergent, 27,
186, 301, 373
set
balanced, 108
bounded, 10, 69, 110
closed, 70, 118
w*-closed, 125, 189
compact, 32, 345, 387, 389, 393
w-compact, 74, 84, 85, 130, 255,
258, 367, 387, 418
w*-compact, 92, 104, 118-, 382
convex, 22, 32, 70, 118, 256, 259,
326
w-closed, 70, 118
w-compact, 84, 279
countably compact, 128, 130, 387
cozero, 393
dense, 14
fragmented, 417
generates space, 357
index, 65
norming, see subspace
orthonormal, 18
point-finite, 393
pointwise bounded, 68
polar (see polar A0 and Ao)
relatively w-compact, 128, 371
saturated family, 119
separated, 30, 54
separating, 93, 134
sequentially closed, 127
sequentially compact, 85, 128, 130,
387, 419
totally bounded, 110
slice, 77
space
angelic, 129, 130, 365, 427
Asplund, 254, 318, 397, 400
B(l2), 15, 233, 235
B(X,Y), 11
Baire, 414, 415
Banach space, 1, 8
c, 6, 14, 56, 99, 100, 153, 154
co, 6, 14, 24, 31, 44, 55, 60, 74,
89, 97, 99, 103, 142, 143,
153-157, 164, 173, 174, 180,
182, 185-187, 193, 195, 205,
233, 267, 269, 274, 276, 280,
291, 306, 331, 343, 345, 355,
358, 375, 377, 380, 381, 383,
401, 415, 421
c0(r), 7, 45, 56, 307, 358, 364, 365,
378, 381, 384, 394
coo j 6
coo(r), 7
C[0,1], 2, 15, 47, 48, 60, 74, 89, 99,
144, 145, 153, 164, 192, 193,
195, 240, 267, 273, 274, 276,
288, 310, 376, 377, 385
Cn[0,l], 24, 154
C[0,wi], 406, 426-428
C°°[0,1], 133

Index 449
space (continued)
C(K), 3, 48, 60, 72, 78, 79, 99, 345,
376, 387, 389, 392, 394, 395,
400, 401, 420, 421
C(Q), 113
r°(Q), 113
complete, 109, 132
complex, 39
(co)type, 30, 306
countable tightness, 129
D, 408
V(?l), 114
Z>'(fi) (see distributions)
direct sum X © Y, 11, 138, 149
direct sum (X 0 Y)Pi 13, 149
direct sum QTXn)p, 56, 271, 306,
381
dual E*, 111
dual ?#, 117
dual X*, 38, 41, 167, 253
dual second X**, 63
T(X,Y), 203, 205
finite-dimensional, 12, 31, 52, 66,
109, 111, 263, 264, 305, 306,
334
finitely represent able, 291, 293,
306, 333
fragmented, 417
Frechet, 108, 129
generated by if, 357
HUbert, 17, 20, 30, 48, 140, 196,
267, 277, 293, 333, 340, 355
homeomorphic, 25, 199, 337, 351,
353
infinite-dimensional, 14, 52, 373
injective, 142, 154
isometric, 11
isomorphic, 11, 307
James J, 185, 198
James tree JT, 199
fC(X,Y), 203, 205
lu 44, 55, 56, 88, 98, 103, 140, 141,
146, 155, 157, 184, 185, 187,
196, 232, 268, 274, 306, 336,
356, 375, 415
?i(r), 56, 156, 267, 358, 367, 400
l2, 15, 20, 21, 31, 88, 89, 98, 155,
177, 193-195, 233, 306, 353,
384, 385
lp, 4-6, 14, 22, 23, 31, 35, 45, 74,
97, 131, 153, 158, 159, 164,
173-175, 180, 193, 195, 205,
306, 336, 341, 342, 385
?P(T), 7, 45, 56, 155, 195, 347
too, 3, 6, 14, 15, 22-24, 44, 62, 88,
95, 99, 141-143, 155, 156, 187,
189, 190, 194, 196, 268, 346,
347, 358, 377, 380, 382-384,
424
*oo(r), 7, 155, 378
C, 306
?CJ, 103, 193, 291, 294
Di[0,l], 22, 47, 99, 183, 276, 358,
385
X2[0,1], 21, 206
ZP[0,1], 8, 15, 22, 23, 45, 74, 112,
154, 177, 180, 193, 268, 302
?oo[0,l], 10, 15, 22, 47, 99, 194, 268
Lp(fji), 10, 286, 290
Lindelof, 404
Lipschitz equivalent, 331, 333
locally convex, 108
metric, 145, 233, 331, 365, 418, 420
metrizable, 71, 72, 110, 390, 391,
393, 399, 421, 428
normable, 108, 111
normed, 1
Polish, 414
pseudocompact, 412
quotient X/Y, 12, 41, 96, 138, 271,
306
real, 39
reflexive, 74, 75, 84, 97, 136, 168,
185, 189, 244, 276, 278, 291,
295, 337, 357, 367, 370, 372,
376, 377, 384
saturated, 343, 345, 401
second dual X**, 69, 82, 167, 275,
291
separable, 14, 26, 41, 55, 71-73, 75,
82, 94, 95, 140-142, 144, 182,
184, 190, 197, 244, 247, 248,
250, 251, 253, 256, 275, 280,
308, 314, 328-330, 337, 345,
348, 353, 357, 377, 391, 410,
412, 425, 428
sequentially compact, 85
sequentially complete, 196

450 Index
space (continued)
Sobolev W*[0,1], 133
superreflexive, 294, 303, 307, 315,
328, 333
topological vector, 107
totally incomparable, 175, 192
Tsirelson T, 311
uniformly homeomorphic, 133
WCD, 383
WCG, 357-359, 364-367, 369, 377,
379-383, 392, 405, 408, 409,
418, 428
weak Asplund, 248
weakly Lindelof, 404, 405, 408-411
span(M), 2, 22
subdifferential, 321
Frechet, 320, 321
subspace, 2, 41, 75
affine, 76
complemented, 137-139, 142,
147-149, 152, 350, 376, 377
finite-dimensional, 13, 109, 139
invariant, 215, 221, 230, 237
norming, 93, 134, 193, 197, 276
proper, 13
quasicomplemented, 377, 379
superdifferential, 320
support supp(/), 114, 328
support supp(x), 5
Sx,2
system, biorthogonal, 139
theorem
Alaoglu, 71, 118
almost block subspace, 173
approximation, 329
Auerbach, 139
Banach contraction, 227
Banach-Dieudonne, 125
Banach-Mazur, 144
Banach-Steinhaus, 68, 69
Banach-Stone, 79
basic sequence, 169, 170, 181
Bessaga-Pelczyriski, 173, 186
bipolar, 119
Bishop-Phelps, 83
Brouwer, 229
Co complementation, 142, 143
Caratheodory, 121
Cauchy-Schwarz, 3, 16
Choquet lemma, 77
Choquet representation, 123
closed graph, 51, 57
DFJP, 366, 371
dual representation, 44, 45, 47, 4
Dvoretzky, 291
Dvoretzky-Rogers, 373
Eberlein-Smulian, 85, 128, 130
extension, 40, 188, 273
Fredholm alternative, 209
Gelfand, 213
Golds tine, 73
Gorelik principle, 353
Hahn-Banach, 37, 40
Helly, 33, 92
Hilbert space is -fe, 20
Holder inequality, 3
invariant subspaces, 230
James, 84, 168, 185
Josefson-Nissenzweig, 88
Kadec, 244, 248, 250, 301
Kaplansky, 129
Khintchine, 177
Korovkin, 240
Krein, 85
Krein-Milman, 76, 101
ti lifting, 141
too injective, 142
local reflexivity, 291, 292
Mackey et al., 120
Markov-Kakutani, 228, 238
Mazur, 70, 118, 422
Milman, 78
Minkowski inequality, 4
norm attained, 40, 83, 262
open mapping, 50, 57
perturbation of basis, 171
Pietsch factorization, 374
Pitt, 175
Ramsey, 423
Riesz lemma, 13
root lemma, 423
Schauder, 207, 229, 230
Schur, 146
selection principle, 173
separable reduction, 254
separable spaces in l\, 140
separable spaces in -ooo)
141, 142

Index 451
theorem (continued)
separable spaces in C[0,1], 144
separation, 41, 43, 69, 70, 80, 117,
330
Simons's inequality, 80
Smulian lemma, 243
Sobczyk, 142
spectral decomposition, 223, 226
subspaces of lp, 174
subspaces of Lp, 177
uniform boundedness, 68, 69
variational principle, 83, 317, 326,
327
topology
locally convex, 108
Mackey, 120, 121
metrizable, 71, 110, 135, 417
norm, 66, 121
(t{E,F), 117
weak w, 64, 96, 129, 391
weak star w*, 64, 71
tree, 199, 294, 295, 308, 309
unit ball Bx, 2
unit sphere Sx, 2
variational principle, 83, 317, 326, 327
vector
cyclic, 237
fixed point, 227
orthogonal, 17, 225, 276
standard unit, 164
weight w(T), 424
Xc, Xn, 39